rnia 

1 


UNIVERSITY  OF  CALIFORNIA 
AT   LOS  ANGELES 


GIFT  OF 

Harriet  B.  Glazier 


&tf 


ELEMENTARY   ALGEBRA 


THE  MACMILLAN  COMPANY 

NEW  YORK   •    BOSTON    •    CHICAGO  •    DALLAS 
ATLANTA   •    SAN    FRANCISCO 

MACMILLAN  &  CO.,  Limited 

LONDON  •    BOMBAY   •    CALCUTTA 
MELBOURNE 

THE  MACMILLAN  CO.  OF  CANADA,  Ltd. 

TORONTO 


Digitized  by  the  Internet  Archive 

in  2007  with  funding  from 

Microsoft  Corporation 


http://www.archive.org/details/elementaryalgebr01cajoiala 


ISAAC  NEWTON 


One  of  the  greatest  mathematicians  of  all  time.  As  a  young  man  he 
invented  the  Binomial  Theorem  which  is  now  studied  in  second  year 
courses  in  algebra.  Newton  wrote  a  Universal  Arithmetic,  which  was 
really  a  book  on  algebra.  He  was  the  first  to  use  fractional  and  negative 
exponents  as  we  write  them.  He  first  used  them  in  a  letter  dated 
June  13,  1676. 


ELEMENTARY  ALGEBRA 


jfirgt  gear  Course 


BY 

FLORIAN    CAJORI 

COLORADO    COLLEGE 
AND 

LETITIA    R.    ODELL 

NORTH    SIDE    HIGH    SCHOOL,    DENVER 


Nefo  gorfe 
THE   MACMILLAN   COMPANY 

1916 

All  rights  reserved 


Copyright,  1915, 
By  THE   MACMILLAN   COMPANY. 

Set  up  and  electrotyped.      Published  June,  1915.      Reprinted 
June,  1916. 


Nortoooli  Iftaaa 

J.  S.  CiiBhIng  Co.  —  Berwick  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


Mathematical 

Sei&nces 

Uorary 


PREFACE 

In  this  book  algebra  is  presented  to  beginners  in  a  simpler, 
clearer,  and  more  practical  form  than  is  usually  found  in 
school  texts. 

The  treatment  is  simplified  by  the  omission  of  certain  re- 
dundant  terms   and   notations,   by   maintaining  an   intimate 
connection  between  algebra  and  arithmetic,  and  by  starting 
with  definite  assumptions  of  the  laws  of  signs  in  subtraction 
O    and  multiplication  rather  than  with  complicated  and  unsat- 
isfactory proofs.     The   part  of  algebra  that  deals  with  the 
P-    mechanical  manipulation  of  algebraic  expressions  is  reduced 
in  amount  as  much  as  is  consistent  with  the  acquirement  of 
■     accuracy.     Simple  fractions  and  easy  radicals  are  introduced 
qO     early ;  the  discussion  of  fractions  with  binomial  and  trinomial 
~7     denominators,  and  consideration  of  the  more  difficult  parts  of 
v     radicals,  are  postponed  to  a  time  near  the  close  of  the  course. 
n,    Ratio  and  proportion  are  brought  into  closer  touch  with  the 
equation  and  with  graphs. 

Clearness  has  been  sought  by  careful  definition,  copious 
illustration,  and   by  the  use  of  language  which  recalls   the 

f  axiomatic  processes  involved.     Such  phrases  as  "  clearing  of 
fractions "  are  objectionable  because  of  the  danger  of  their 
,     being  applied  by  the  pupil,  not  to  the  equation  alone,  but  to 
J     algebraic  expressions  in  general.     "  Canceling,"  as  now  used, 
^  is  ambiguous,  for  it  sometimes  involves  subtraction,  at  other 
gh     times  division. 

The  treatment  is  rendered  more  practical  by  a  careful  selec- 
tion of  problems.     Mechanical  problems  involving  the  theory 


348554 


Vi  PREFACE 

of  the  lever,  specific  mass,  specific  heat,  the  motion  of  falling 
bodies,  and  technical  terms  in  electricity  are  omitted  at  first, 
and  even  near  the  end  they  are  used  only  sparingly.  Expe- 
rience has  shown  that  first  year  pupils  in  the  high  school 
cannot  cope  successfully  with  the  abstract  concepts  of  me- 
chanics. These  topics  belong  more  properly  to  the  course  in 
physics  usually  given  in  the  third  or  fourth  year.  In  this 
text  stress  is  laid  on  practical  applications  to  problems  arising 
in  business.  In  such  applications  the  pupil  is  merely  continu- 
ing on  a  somewhat  higher  plane  the  subjects  first  approached 
in  arithmetic.  A  distinguishing  feature  of  this  text  is  the 
practical  application  of  graphs.  The  ordinary  procedure  is 
to  use  the  graph  merely  in  presenting  to  the  eye  the  behavior 
of  two  variables  in  an  equation.  In  this  text  it  serves,  in 
addition,  for  the  determination  by  inspection  of  results  de- 
cidedly practical  in  character. 

The  authors  are  indebted  for  valuable  criticisms  and  sug- 
gestions to  several  teachers,  but  more  particularly  to  Principal 
E.  L.  Brown  of  the  North  Side  High  School  in  Denver. 

FLORIAN  CAJORI. 
LETITIA   R.    ODELL. 


CONTENTS 


I.  Introduction         ...... 

II.  Positive  and  Negative  Numbers 

III.  Multiplication  and  Division 

IV.  Proportion,  Graphs     ..... 
V.  Equations  involving  Fractional  Coefficients 

VI.  Special  Products  ..... 

VII.  Factoring 

VIII.  Equations  solved  bt  Factoring 

IX.  Multiplication  and  Division  of  Polynomials 

X.  Square  Root 

XI.  Graphs  of  Simultaneous  Equations  . 

XII.  Simultaneous  Linear  Equations 

XIII.  Quadratic  Equations  ..... 

XIV.  Fractions 

XV.  Radicals  and  General  Exponents    . 

XVI.  Review  of  Essentials  of  Algebra  . 


1 

14 

41 

58 

77 

84 

89 

104 

111 

114 

119 

126 

141 

162 

178 

192 


vii 


Q    -r   T- 


cie 


ELEMENTARY  ALGEBRA 


CHAPTER   I 

INTRODUCTION 

1.  Algebra,  like  arithmetic,  deals  largely  with  the  solution 
of  problems  involving  numbers.  It  is  an  extension  of  arith- 
metic which  enables  one  to  solve  more  complicated  problems. 
It  extends  the  field  of  arithmetic  by  means  of  three  devices : 
the  use  of  letters  as  well  as  Hindu- Arabic  numerals  in  the 
study  of  numbers ;  the  introduction  of  new  kinds  of  numbers 
to  be  used  in  conjunction  with  those  of  arithmetic ;  and  the 
development  of  simple  methods  of  operation  with  those 
numbers. 

The  signs  of  operation  used  in  arithmetic  are  used  also  in 
algebra. 

Thus  addition  is  indicated  by  -f,  subtraction  by  — ,  multiplication  by 
X ,  division  by  -f- . 

The  sign  of  multiplication  (  x  )  is  often  replaced  in  algebra  by  a  dot  (•), 
which  is  written  above  the  line  to  distinguish  it  from  a  decimal  point. 
Thus  3  •  4  means  3x4. 

The  product  of  two  numbers,  one  or  both  of  which  are*  represented  by 
letters,  is  usually  indicated  by  writing  the  numbers  one  after  the  other 
without  any  sign  between  them.  Thus  ab  means  a  x  b  or  a  •  6,  4  c  means 
4  x  c  or  4  •  c. 

When  only  one  of  two  numbers  is  represented  by  a  letter,  it  is  custom- 
ary to  write  the  letter  last.     Thus  we  write  4  c,  but  not  c  4. 

In  arithmetic,  and  also  in  algebra,  division  is  frequently  expressed  in 
the  form  of  a  fraction.     Thus  -  =  4  -i-  5,  -  =  a  -4-  6. 

=  is  the  sign  of  equality. 

2.  Any  combination  of  numbers,  letters,  and  symbols  of  opera- 
tion, which  represents  a  number,  is  called  an  algebraic  expression. 

Thus,  a  +  6,  a~    ,  4 ab  +  c  are  algebraic  expressions. 
2  c 


2  ELEMENTARY  ALGEBRA 

3.  In  arithmetic  it  is  customary  to  use  abbreviations,  such 
as  ft.  for  "foot,"  in.  for  "inch,"  A.  for  "acres," and  so  on.  In 
algebra  the  practice  of  using  abbreviations  is  carried  much 
further.  If  we  write  i  for  "  interest,"  p  for  "  principal,"  r  for 
"  rate,"  and  t  for  "  time,"  then  the  statement 

"  The  interest  is  equal   to  the  product  of  the  principal,  the 

rate,  and  the  time  " 
can  be  expressed  briefly  by  the  equation 

i  =  prt. 

By  the  use  of  such  a  system  of  shorthand  much  time  and 
labor  is  saved.  The  same  letter  may  be  used  for  different 
things  in  different  problems. 

The  letter  m  may  be  used  for  "  miles  "  in  one  problem  and 
"minutes"  in  another.  The  letter  x  is  used  extensively  to 
designate  a  number  which  is  unknown  at  the  outset,  but  which 
is  to  be  determined  by  the  solution  of  the  problem  in  hand. 
In  one  case  x  may  mean  the  number  of  "  dollars,"  in  another 
case  the  number  of  "  pounds,"  in  a  third  case  the  number  of 
"  years."  The  meaning  of  a  letter  must  be  made  plain  in  each 
problem.  To  avoid  confusion,  no  letter  should  stand  for  two 
different  things  in  the  same  problem. 

If  i  means  "  inches  "  and  /  means  "  feet,"  then 
.48  i  =  4/  means  "  48  inches  are  equal  to  4  feet." 

15/+  4  i  means  "  15  feet  and  4  inches."  If  n  stands  for  any 
number,  then  2  n  -f  50  means  "  2  times  any  number,  increased 
by  50,"  or  "  twice  any  number,  plus  50."  If  n  is  taken  equal  to 
3,  then  In  +  50  =  2  •  3  +  50  =  56.  If  n  =  25,  then  2n  +  50 
=  2  •  25  +  50  =  100. 

ORAL   EXERCISES 

4.  Express  in  algebraic  symbols  a  number  which  is 

1.  5  more  than  x.  4.   10  less  than  five  times  c. 

2.  5  less  than  x.  5.   20  less  x. 

3.  .8  more  than  three  times  x.        6.   1.5  less  five  times  x. 


INTRODUCTION  3 

7.  6  less  than  y.  9.    .18  less  than  ten  times  z. 

8.  1.2  more  than  y.  10.    5  more  than  \  of  y. 

11.    1  less  than  twice  y. 

Supplying  for  each  letter  a  given  number,  express  in  words 
exercises  12-25 : 


12. 

16  g. 

19. 

i2/  +  7. 

13. 

g  +  i- 

20. 

3y  +  5. 

14. 

2z+31. 

21. 

7-4y. 

15. 

2x-5. 

22. 

4y-iy-5. 

16. 

7  x  -  13. 

23. 

15 --§</. 

17. 

13  —  5  #. 

24. 

17  x  +  5  x  =  22  x. 

18. 

i2/-3. 

25. 

12*8—3?  =  9a. 

26. 

If  i  =  interest,  p 

=  principal, 

y  =  time,  r  =  rate,  express 

in  words  i  —  prt,  p  =  —,  r  =  — ,  £  =  — . 
r£  pt  '      pr 

Find  the  values  of  the  following  expressions  if  a  =  2,  6  =  3, 
c  =  5,  d  =  19 : 

27.  8a +  26.  32.  a+26  +  3c-d. 

28.  3  6  +  a  33.  d  —  5  a  +  7  6  —  c. 

29.  6c- 6.  34.  ^  + a. 

30.  9c  +  4d.  35.  59  — 5c +  10 6. 

31.  a  +  6  +  c  +  d.  36.  2d  +  199c  -25a  -19. 

If  m  =  15,  n  =  12,  p  =  20,  q  =  0,  what  are  the  values  of  the 
following  algebraical  expressions? 

37.  m  +  n+p  +  q.  ■       39.    12n  +  2p  +  5m  —  q. 

38.  q  +  5p  —  6m.  40.   6p  +  12  +  2n  —  3 m. 
(When  g=0,  then  12  g  is  equal  to  12  times  0 ;  this  gives  the  product  0.) 

41.  2  m  +  r  +  12  a.  43.    5  n  +  m  —  7  q. 

42.  3p  +  5  q  +  m.  44.    8  g  +  5 p  —  2  m. 


ELEMENTARY  ALGEBRA 


Fig.  1 


EQUATIONS 

5.  An  equation  expresses  an  equality.  In  other  words,  an 
equation  is  a  statement  that  two  expressions  stand  for  the 
same  number.  Thus  x  +  3  =  3x-  1  is  an  equation.  Equa- 
tions  are  used  in  the  solution   of   problems.     Some  number, 

which  at  the  outset  is  un- 
known, is  represented  by 
a  letter;  an  equation  is 
formed,  which  enables  one 
to  ascertain  the  value  of 
that  unknown  number. 

An    equation    is    like   a 
balance   which  is  in   equi- 
librium when  the  weights 
placed    in    one    scale    pan 
are  together  equal  to  the 
weights  placed  in  the  other 
scale  pan. 
The  equilibrium  of   a   balance  is  not  disturbed  so  long  as 
like  changes  in  the  weights  are  made  simultaneously  on  both 
sides. 

So  in  equations,  we  may  add  the  same  number  to  both  sides, 
or  subtract  the  same  from  both  sides,  or  we  may  multiply  or 
divide  both  sides  by  the  same  number  (except  division  by 
zero) ;  the  equality  is  maintained  during  all  these  changes. 

On  the  other  hand,  the  equality  is  destroyed  if  more  is 
added  to  or  subtracted  from  one  side  than  the  other,  or  if  one 
side  is  multiplied  or  divided  by  a  larger  number  than  is  the 
other  side. 

6.  Illustrative  Problem.  If  13  is  added  to  twice  a  cer- 
tain number,  the  sum  is  41.     Find  the  number. 

Arithmetical  Solution.  Here  we  do  not  use  equations.  We  subtract  13 
from  41  and  obtain  28.  Then  we  divide  28  by  2  and  obtain  14,  which  is  the 
required  number.    That  is,  we  begin  with  41,  and  go  back  to  the  required 


INTRODUCTION  5 

number  by  subtraction  and  division,  which  are  respectively  the  inverse  of 
the  operations  of  addition  and  multiplication,  named  in  the  problem. 

Algebraical  Solution.  Here  we  use  equations.  Let  some  letter  stand 
for  the  unknown  number  and  then  perform  upon  it  the  direct  operations 
named  in  the  problem  and  obtain  an  equation.  This  direct  way  is  usually 
easier  than  the  inverse.     The  solution  is  as  follows  : 

Let  x  be  the  unknown  number.  Then  2  x  is  twice  that  number,  and 
2  x  +  13  is  the  sum  obtained  by  adding  13  to  twice  that  number. 

But  this  sum  is  equal  to  41,  as  is  stated  in  the   problem.     That  is, 

2  x  +  13  =  41. 

We  want  to  find  x.  Remembering  that  an  equality  is  like  a  balance, 
that  the  equality  is  not  destroyed  if  13  is  subtracted  from  both  sides,  we 

obtain  2  x  +  13  -  13  =  41  -  13,  or 

2  x  =  28. 

The  equality  remains  true  if  both  sides  are  divided  by  2.     Hence 

2x      28 
—  =  — ,  or 
2        2 

x  =  14. 

We  have  now  solved  the  equation  2  x  +  13  =  41  and  have  found  the 
previously  unknown  number  x  to  be  equal  to  14.  We  see  that  the  solu- 
tion of  the  equation  consists  in  making  such  changes  in  the  equation  that 
x  finally  stands  alone  on  one  side  of  the  equation. 

7.  Second  Illustrative  Problem.  If  from  ^  of  a  certain 
number  17  is  subtracted,  the  number  resulting  is  35.  Find 
the  number. 

Let  x  be  the  required  number.     Then 

x      1 

-  is  -  of  that  number,  and 

3      3 

x  1 

17  is  -  of  that  number,  less  17. 

3  3 

But  this  difference  is  equal  to  35,  as  is  stated  in  the  problem.     Hence 

-  —  17  =  35.     To  solve  this,  add  17  to  both  sides.     We  get 

o 

-  =  52.     Multiplying  both  sides  by  3,  we  obtain 

o 

x  =  156,  which  is  the  answer. 

This  answer  is  correct,  for  £  of  156  =  52,  and  52  —  17  =  35,  as  stipu- 
lated in  the  problem. 


6  ELEMENTARY  ALGEBRA 


PROBLEMS 


8.  1.  If  23  is  added  to  three  times  a  certain  number,  the  sum 
is  74.     Find  the  number. 

2.  A  wagon  loaded  with  12  sacks  of  flour  weighs  1876  lb.; 
the  empty  wagon  weighs  700  lb.     Find  the  weight  of  1  sack. 

3.  A  train  travels  from  Baltimore  to  New  York,  a  distance 
of  185  mi.,  in  4  hr.     Find  the  average  velocity  of  the  train. 

Work  by  arithmetic ;  then  by  algebra. 

In  the  latter  case,  let  v  be  the  velocity  in  miles  per  hour ;  form  an 
equation  which  expresses  the  relation  : 

velocity  x  time  =  distance. 

4.  The  distance  from  New  York  to  St.  Louis  is  1058  mi. 
What  is  the  velocity  of  a  train  which  travels  this  distance  in 
23  hr.  ? 

5.  How  long  will  it  take  a  train  to  travel  the  distance  of 
960  mi.  from  New  York  to  Chicago  at  an  average  speed  of 
48  mi.  an  hour? 

Let  t  =  the  no.  of  hours  ("  time  "). 

6.  A  lot  sold  for  $2475.  What  was  the  frontage,  if  it  sold 
at  $  75  a  front  foot  ? 

7.  The  sura  of  two  numbers  is  31 ;  the  larger  exceeds  the 
smaller  by  7.     Find  the  two  numbers. 

Let  the  smaller  number  =  x. 

Then  the  larger  number  =  x  +  7. 

The  sum  of  the  two  numbers       =  x  +  x  +  7. 

But  the  sum  of  the  two  numbers  =  31. 

Hence  x  +  x  +  7  =  31. 

Since  a-  -f  a:  =  2  x,  2z  +  7  =  31. 

Subtract  7  from  both  sides,     2  x  =  24. 

Divide  both  sides  by  2,  x  =  12,  the  smaller  number. 

Whence  x  +  7  =  19,  the  larger  number. 

8.  The  sum  of  two  numbers  is  75;  the  larger  exceeds  the 
smaller  by  15.     Find  the  two  numbers. 

9.  Find  two  numbers  whose  difference  is  34  aud  whose  sum 
is  126. 


V 


INTRODUCTION  7 

10.  A  father  earns  S  24  a  week  more  than  his  son.  Together 
they  earn  $  36  a  week.      What  are  the  weekly  wages  of  each  ? 

11.  Divide  $  96  between  father  and  son  so  that  the  son  gets 
|  of  what  the  father  gets. 

12.  Divide  $  124  between  two  brothers  so  that  one  receives 
$  36  more  than  the  other. 

13.  I  propose  to  a  boy  the  following  puzzle :  "  Think  of  a 
number,  multiply  it  by  10,  add  30,  subtract  20."  He  gives 
the  result  as  60.     Find  the  number. 

14.  To  find  the  weight  of  a  golf  ball  a  man  puts  10  golf 
balls  into  the  left  scale  pan  of  a  balance  and  a  1-lb.  weight  into 
the  right ;  he  finds  that  too  much,  but  the  balance  is  restored 
if  he  puts  1  oz.  into  the  left  scale  pan.  What  was  the  weight 
of  a  golf  ball  ? 

15.  A  path  half  a  mile  long  is  to  have  a  curb  on  both  sides. 
V/The  curbstones  used  are  40  in.  long,  and  200  of  them  have 

already  been  supplied.     How  many  more  are  wanted  ? 

16.  The  cost  of  housekeeping  for  a  family  of  n  persons  is 
estimated  at  7.5  +  2.5  n  dollars  per  week.  How  many  persons 
are  there  in  a  family  whose  weekly  expenses  are  $  25  ? 

17.  A  father  leaves  $  14,000  to  be  divided  among  his  three 
children,  so  that  the  eldest  child  receives  $  1000  more  than  the 
second,  and  twice  as  much  as  the  third.  What  is  the  share  of 
each? 

18.  Divide  $24,000  among  A,  B,  and  C,  so  that  A's  share 
may  be  three  times  that  of  B,  and  C  may  have  £  of  what  A  and 
B  have  together. 

-  19.  A  piece  of  silver  is  found  to  weigh  a  certain  number  of 
ounces  in  a  pair  of  scales ;  on  taking  out  a  weight  of  3  oz.  from 
one  scale  and  placing  it  in  the  scale  containing  the  silver,  the 
contents  of  one  scale  is  twice  as  heavy  as  the  contents  of  the 
other.     Determine  the  weight  of  the  silver. 


8  ELEMENTARY  ALGEBRA 

20.  Find  the  number  whose  excess  over  2  is  three  times  the 
excess  of  its  half  over  4. 

21.  If  the  length  of  a  circle  is  3.1416  times  its  diameter, 
find  the  radius  of  a  circle  (correct  to  two  decimal  places),  when 
the  length  of  the  circle  is  7.56  ft. 

To  determine  the  nearest  figure  in  the  second  decimal  place,  it  is 
necessary  to  compute  the  figure  in  the  third  decimal  place.  If  the  deci- 
mal figures  are,  for  example,  .546,  write  .55 ;  if  the  decimal  figures  are 
.543,  write  .54 ;  if  the  decimal  figures  are  .545,  write  either  .54  or  .65. 

22.  If  6  be  added  to  7  times  a  certain  number,  the  result  is 
equal  to  144.     Find  the  number,  correct  to  two  decimal  places. 

23.  The  sum  of  two  numbers  is  11.3;  8  times  the  greater 
exceeds  11  times  the  smaller  number  by  2.5.  Find  the  num- 
bers, correct  to  three  decimal  places. 

FACTOR,   COEFFICIENT,  TERM,  EXPONENT 

9.     Each  of  the  quantities  which  multiplied  together  form 
a  product  is  called  a,  factor  of  the  product. 

Thus,  each  of  the  numbers  3,  5,  x,  y,  is  a  factor  of  the  product  15  xy. 
Also,  since  15  x  times  y  gives  the  product  15  xy,  15  x  is  called  a  factor  of 

15  xy. 

In  general,  the  product  of  any  two  of  the  simple  factors, 
3,  5,  x,  y,  is  also  called  a  factor  of  15 xy;  also,  the  product  of 
any  three  of  the  factors  3,  5,  .r,  y,  is  called  a  factor  of  15  xy. 

10.    In  4  c,  4  is  called  the  coefficient  of  c.     Similarly, 

In  16  a,  16  is  called  the  coefficient  of  a. 

In  16  ab,  16  is  called  the  coefficient  of  ab. 

Here  the  word  coefficient  is  applied  to  the  factor  which  is 
expressed  in  the  Hindu-Arabic  numerals.  This  is  the  most 
common  use  of  the  word  coefficient,  but  in  a  broader  sense, 
either  of  two  factors  which  are  multiplied  together  to  form  a 
product  is  called  a  coefficient  of  the  other  factor. 

For  instance,  16  ab  is  the  product  of  two  factors  16  a  and  b  ;  hence 

16  a  is  the  coefficient  of  b,  and  b  is  the  coefficient  of  16  a. 


INTRODUCTION  9 

If  no  numerical  coefficient  is  expressed,  1  is  understood ;  x 
is  the  same  as  1  x.     Notice  also  that  0  x,  or  0  x  x,  is  equal  to  0. 

11.  An  algebraic  expression  may  consist  of  parts  which  are 
separated  by  the  +  or  —  signs ;  these  parts  with  the  signs 
immediately  preceding  them  are  called  terms. 

Thus,  the  expression  3o  —  46  +  5c  is  separated  by  the  +  or  —  signs 
into  three  parts ;  it  has  the  terms  +  3  a,  —  4  6,  +  5  c.   / 

12.  An  algebraic  expression  of  one  terra  is  called  a  monomial, 
of  two  terms  a  binomial,  of  three  terms  a  trinomial,  and  of 
several  terms  &  polynomial. 

13.  The  product  of  two  equal  factors  a  •  a  is  called  the  square 
of  a. 

a  ■  a  is  usually  written  a2  and  is  read  "  a  square  "  or  "  the  second 
power  of  a." 

The  product  of  three  equal  factors  a  ■  a  •  a  is  called  the  cube  of  a. 

a-aa  is  usually  written  a3,  and  is  read  "a  cube"  or  "the  third 
power  of  a." 

The  product  of  four  equal  factors  a  •  a  •  a  •  a  is  called  "  a  fourth1'  or 
"  the  fourth  power  of  a." 

In  a2,  a3,  a4,  a5,  a  is  called  the  base,  the  2,  3,  4,  5,  are  called  exponents. 

14.  An  exponent  is  a  number  or  letter  placed  a  little  above 
and  to  the  right  of  another  number  or  letter,  called  the  base. 

When  the  exponent  is  a  positive  integer,  it  indicates  how 
many  times  the  base  is  taken  as  a  factor. 

Avoid  saying  that  a2  means  a  multiplied  by  itself  2  times.  This  is  not 
true,     a'2  means  a  •  a,  where  the  base  a  is  multiplied  by  itself  only  once. 

Why  is  it  incorrect  to  say  that  a8  means  a  multiplied  by  itself  3  times  ? 

When  no  exponent  is  expressed,  the  exponent  is  regarded  as  1.  Thus, 
a  means  the  same  as  a1,  7  the  same  as  71. 

Later  we  shall  extend  the  definition  of  an  exponent,  in  order  to  give 
meaning  to  such  expressions  as  5c*,  or  c5. 

Care  must  be  taken  to  distinguish  between  exponents  and  coefficients. 
Notice  that  e4  means  e  x  e  x  e  x  e,  but  4  e  means  e  +  e  ■{-  e  +  e.  Ife  = 
3,  then  e4  =  3  •  3  •  3  •  3  =  81,  while  4e  =  3  +  3  +  3  +  3=12. 


10  ELEMENTARY  ALGEBRA 

15.  If  the  factors  of  a  number  are  all  equal,  any  one  of 
them  is  called  a  root  of  the  number. 

Nine  has  two  equal  factors,  3  and  3.  We  call  3  the  square  root  of  9. 
Similarly,  27  has  three  equal  factors  3,  3,  and  3 ;  3  is  called  the  cube 
root  of  27.  Again  256  has  four  equal  factors,  each  being  4;  hence  4  is 
called  the  fourth  root  of  256. 

Roots  are  indicated  as  follows:  ^9  =  3,  ^27=3,  ^256  =  4.  The 
figure  of  the  radical  sign  shows  what  root  of  the  number  is  to  be  taken. 
This  figure  is  called  the  index  of  the  root.  If  no  figure  is  expressed, 
square  root  is  understood. 

EXERCISES 

16.  Express  in  words  the  following : 

1.  a5.  6.  41m.  11.  ^125-^64. 

2.  63.  7.  3y2.  12.  193-54. 

3.  a5  +  &3.  8.  5fc*  +  c.  13.  a  +  s2  -f  g5  —  8p7. 

4.  e2  — 4.  9.  6/i  —  9x.  14.  x2  +  y2  +  z  —  mn. 

5.  82-29.  10.  V36-V25.  15.  h% -u* +  W-y/Sx. 

16.  Compute  the  value  of  the  expression  in  exercises  1-15, 
if  a  =  2,  6  =  4,  c  =  50,  e  =  30,  g=l,  h  =  5,  fc  =  ll,  ra  =  6, 
w  =  10,  p  =  l,  s  =  12,  u  =  0,  x  =  S,  y  =  9,  z  =  29. 

17.  Find  the  value  of  4  c4  for  each  of  the  following  values 
if  c  =  1,  2,  5,  6,  4,  3,  10. 

18.  When  s  =  10,  compute  the  values  of  s2  — 2  s,  s3  —  3  s, 
s4  —  4  s,  s5  —  5  s. 

19.  When  t  =  |,  compute  the  values  of  f2,  f3,  5  P  +  6  ts,  3  t  - 1?, 
At-t*. 

20.  When  b  =  0,  find  the  values  of  &2,  b\ 

PARENTHESES 

17.  When  terms  are  to  be  grouped  together,  parentheses  are 
used. 

Thus,  (x  +  y)2  means  that  x  +  y,  considered  as  a  single  num- 
ber, is  to  be  squared.     That  is,  (x  +  y)2  =  {x-\-y)  (x  +  y). 


INTRODUCTION  11 

Again,  5a—  (b  —  2c-\-d)  means  that  the  entire  expression 
b  —  2  c  +  d  is  to  be  subtracted  from  5  a. 

When  several  parentheses  are  used  in  the  same  expression, 
confusion  may  be  avoided  by  using  different  forms.  All  these 
forms  go  by  the  general  name  of  "  parentheses,"  but  they  are 
designated  by  special  names  when  it  is  desirable  to  distinguish 
between  them.     Thus  [  ]  is  called  a  "  bracket,"  \  \  is  called  a 

"  brace," is  called  a  "  vinculum."     But  ( )  is  always  called 

a  "  parenthesis." 

That  the  sum  of  a  and  b  is  to  be  multiplied  by  c  may  be  indicated  in 
four  different  ways  as  follows  :  (a  +  b)c,  [a  +  6]c,  {a  +  b}c,  a  +  b  ■  c. 

EXERCISES 
18.    Read  and  tell  the  meaning  of  expressions  1-12 : 


1. 

20(5  -  3). 

8, 

(2-s)  +  [10-4]. 

2. 
3. 

10(*-22f). 

(x+y)(x-y). 

9. 

x      \5     3) 

4. 
5. 
6. 

(±x+3)(x-y). 
3  x(x  +  y). 
3x+(x  +  y). 

10. 
11. 

b  +  c    d. 
e+f-\g  +  h\. 

7. 

(c  +  d)-5. 

12. 

[a  +  6-(c4-d)] 

13.  Showthata2  —  b2  =  (a  -  6)(a+  b),  whena  =  12and6  =  10. 

14.  Showthat  \a  +  b\2  =  a?  +  b*  +  2 ab,  whena  =  9and6  =  5. 

15.  Show  that  (a  -  bf  =  a2  +  b2-2  ab,  when  a  =  12,  6  =  8. 

19.   Expressions  like  19  4-  (7  +  3)  may  be   worked  out   in 
two  ways: 

(1)  First  simplify  inside  the  parenthesis  and  then  add. 

(2)  Add  7  to  19  and  then  add  3  to  the  result.     That  is, 

19 +(7 +  3)=  19 +  10  =  29, 

'=19  +  7  +  3  =  26  +  3  =  29. 


12  ELEMENTARY  ALGEBRA 

In  the  same  way  there  are  two  ways  of  working  19  +(7  —  3) 
or  19  -  (7  +  3)  or  9  x  (7  +  3)  or  9  X  (7  —  3),  as  appears  from 
the  following : 

(1)  19 +(7 -3)=  19 +  4  =  23, 

(2)  =  19  +  7  -  3  =  23. 

(1)  19 -(7 +  3)=  19 -10  =  9, 

(2)  =  19  -  7  -  3  =  9. 

(1)  9x(7+3)=9xl0  =  90, 

(2)  =9x7  +  9x3  =  90. 

(1)  9  x(7-3)=  9x4=36, 

(2)  =  9  x  7  -  9  x  3  =  36. 

EXERCISES 

20.  Perform  each  of  the  following  exercises  in  two  ways : 

1.  8 +[6 +  3].  6.  5x(5  +  5). 

2.  |10  +  5|  +  122.  7  [3  +  9+5-3]x5. 

3.  30 -[4 +  8]. 

4.  16+(ll-6).  8-  ^x(l  +  2  +  3-6). 

5.  15 +[5 +6- 2].  9-  30(10-4  +  3-1). 

21.  In  expressions  like  9-4  +  3  or  5-6-2- 10 +  23  or 
3  +  3-6-5-3-2  it  is  understood  that  the  multiplications  and 
divisions  (if  there  are  any)  are  performed  first  in  their  order 
from  left  to  right;  the  additions  and  subtractions  are  carried 
out  afterwards  in  their  order  from  left  to  right. 

Notice  that,  in  a  term  like  3  •  5  -s-  3  •  2,  it  is  understood  that 
the  divisor  is  3,  not  3*2.  If  we  want  3  •  2  to  be  the  divisor, 
we  must  inclose  it  in  a  parenthesis  and  write  the  expression 
thus,  3  •  5  -f-  (3  •  2).  Then  3  ■  2  will  be  regarded  as  a  single 
number.     Bearing  these  things  in  mind,  we  see  that 

5-4  +  3  =  1  +  3  =  4. 

5-6-2-  10  +  2-3  =  30-20  +  6  =  16. 

3  +  3-6-^-3-2  =  3  +  18  -=-3-2  =  3  +  6-2  =  15. 

3  +  3-6-=-(3-2)  =  3  +  18h-6  =  3  +  3  =  6. 


INTRODUCTION 


13 


ORAL.  EXERCISES 

22.  Simplify  the  following : 

1.  10-5  +  3-8. 

2.  10 -(5 +  3)  + 8. 

3.  10x2-9-3  +  5x2. 

4.  12  -4-5-2  -4-3. 


5.  12-4-*-(2-4)-3. 

6.  12-4  +  2(4  —  3). 

7.  10 -[12 -3 -2]. 

8.  10 -12 -3 -2. 


Of  the  following  results,  which  are  wrong  ? 
9.   50-10  +  8  =  50-18  =  32. 

10.  20  +  24  -=-  3  •  4-2  =  20  +  24  - 12  -  2  =  20  +  2  -2  =  20. 

11.  15  +  36  -=-  (3  +  9)  +  5  =  15  +  36  -r-  3  +  9  +  5  =  41. 


EVALUATION   OF  ALGEBRAIC  EXPRESSIONS 

23.    When  a  =  5,  6  =  3,  c  =  2,   d  =  6,   and   e  =  4,   find  the 

values  of : 


i.  M- 

b      c 

2.  a&c  +  d2-e3. 

3.  5a3-(2a6  +  d). 
d-6 


5. 


10a  +  86-e 


G. 


a2  +  b3  -  3  6 


c  +  d2  -  a2  +  7 

7.  2fa  +  e-d  +  64|-46. 

8.  (a  +  b){c  x  d  —  5). 

9.  6V36-cVl6  +  dVl44. 
10.    Va+&  +  c  +  d  +  e  — 4. 


a  +  c+  e 

11.  Does  a2 +  5 a  —  36  =  0,  when  x  =  4?     Whenx  =  5? 

12.  Does  x3  -  x2  +  x  -  10  =  18,  when  x  =  5  ?     When  x  =  3  ? 


CHAPTER  II 

POSITIVE  AND   NEGATIVE  NUMBERS 

24.    The  thermometer  in   your  classroom  indicates  a  tem- 
perature of,  perhaps,  "67°  above  zero."     In  the  northern  states 
the  winter  temperature  out  of  doors  sometimes  drops 
F-      to  "  15°  below  zero  "  or  even  lower.     A  shorter  way 
2l2tf212     °^  expressing  this  is  as  follows : 

For  "67°  above  zero"  write  "  +67°." 
For  "  15°  below  zero"  write  "  -  15°." 
A  -f  indicates  that  the  temperature  is  "  above  zero"  ; 
-100°    a  —  indicates  that  it  is  "below  zero." 

But  such  plus  and  minus  numbers  are  convenient 
in  other  ways.  A  man  who  takes  in  and  pays  out 
money  may,  for  brevity,  mark  the  sums  taken  in  by 
prefixing  the  +  sign,  and  the  sums  paid  out  by  pre- 
:  -32*  fixing  the  —  sign.  It  is  customary  also  to  indicate 
the  amount  a  man  owns  by  +  and  the  amount  a 
man  owes  by  — . 

In  the  same  way  we  may  write  1916  a.d.  as  +1916 
-10°    -10°    an(^  ^^  BC#  as  —  ^k     These  illustrations  make  it 
-20°    -20°    plain  how  plus  and  minus  numbers,  or  positive  and 
negative  numbers,  as  they  are  more  usually  called, 
^         may  be  used  in  ordinary  affairs  of  life.     As  we  pro- 
Fig.  2      cee(l  further  we  shall  see  that  the  use  of  such  num- 
bers makes  the   solution  of  many   problems  much 
easier  and  shorter.     Such  numbers  can  always  be  used  when 
there  are  pairs  of  quantities  which  are  the  exact  opposites  of 
each  other,  as  are  the  quantities  in  the  above  illustrations. 

14 


POSITIVE  AND  NEGATIVE  NUMBERS  15 

ORAL  EXERCISES 

25.  State  the  answers  by  prefixing  +  or  —  to  the  numbers : 

1.  The  temperature  at  noon  is  +  40°  and  falls  25°  by  night. 
State  the  temperature  at  night.  Give  the  answer  if  it  falls  50° 
(that  is,  if  it  falls  40°  and  then  10°  more). 

2.  A  boy  purchases  a  book  which  costs  $  1.75  and  pays  $  .50 ; 
the  balance  he  has  charged.  How  may  he  indicate  the  amount 
charged  ? 

3.  His  father  gives  him  two  dollars  and  tells  him  to  pay  the 
balance  on  that  book.     How  much  has  the  boy  left  ? 

26.  One  of  the  most  common  modes  of  representing  positive 
and  negative  numbers  to  the  eye  is  by  distances  along  a  straight 
line,  as  is  done  in  the  thermometer.  It  does  not  matter  in 
what  direction  the  line  is  drawn.  Usually  it  is  most  conven- 
ient to  draw  the  line  horizontally, 

Some  point  0  is  taken  as  the  starting  point  (corresponding  to 
the  zero  in  the  thermometer).  Distances  to  the  right  of  0  are 
usually  indicated  by  the  positive  numbers,  and  distances  to  the 
left  of  0  by  the  negative  numbers. 

I   i    i    i    I    i    i    >    [    i   ■    i    i    i    i    '    I   i   i    »   I  '    i    ■   I   '    i    '    I   i   i   i   I 
-4  -3  -2  -1  0  +1  +2  +3  +4 

Fig.  3 

In  the  figure,  the  distance  from  0  to  + 1  is  taken  as  unit 
distance  or  1  space.  The  point  marked  +  2  is  2  spaces  to  the 
right  of  0;  the  point  marked  —2  is  2  spaces  to  the  left  of 
0. 

In  the  same  way,  a  number  +  2\  is  indicated  by  a  point  2\ 
spaces  to  the  right  of  0 ;  —  2~\  is  indicated  by  a  point  2\  spaces 
to  the  left  of  0. 

If  to  +1  we  desire  to  add  +  2,  we  start  at  the  point  marked 
+ 1,  and  go  2  units  to  the  right,  giving  +  3  as  the  answer. 

If  from  + 1  we  desire  to  subtract  +  2,  we  start  at  +  1  and  go 
2  units  to  the  left,  giving  —  1  as  the  answer. 


16  ELEMENTARY  ALGEBRA 


ADDITION 
<C  ORAL  EXERCISES 

27.  1.    Locate  on  the  line  (Fig.  3)  the  following  numbers  : 

+  4,-3,  +1|,  -2|,3i,i. 

2.  To  -  1  add  3. 

We  start  at  the  point  —  1,  and  add  3  by  going  3  units  to  the  right, 
giving  +  2  as  the  answer. 

3.  From  + 1  subtract  4. 

4.  Add  4  and  —3,-4  and  -  3,  + 1|  and  2,  +  4|  and  -  2. 

5.  Add  +5^  and  —  2,  +4£  and  1£,  +  6  and  -5, 15  and  —  25. 

28.  The  student  will  have  noticed  that  each  of  the  symbols 
+  and  —  is  used  in  algebra  in  two  senses.  Thus  +  is  used 
sometimes  to  indicate  addition  and  at  other  times  it  is  used 
to  show  that  the  number  to  which  it  is  prefixed  is  positive. 

Similarly,  —  sometimes  signifies  subtraction  and  at  other 
times  is  used  to  represent  a  negative  number.  This  double  use 
of  these  symbols  may  at  first  seem  confusing,  but  we  soon  learn 
how  to  interpret  algebraic  expressions  in  which  they  are  vised. 

To  express  in  algebraic  symbols  the  addition  of  +  5  and  —4 
we  inclose  the  numbers  in  parentheses  and  write  thus : 

(+5)  +  (-4). 
The  +  between  the  two  parentheses  means  addition;  the  + 
in  (+5)  and  the  —  in  (  —  4)  indicate  whether  the  number  is 
positive  or  negative. 

29.  By  the  absolute  value  of  a  number  is  meant  its  value 
without  regard  to  the  sign  before  it.  Thus  the  absolute  value 
of  both  +  5  and  —  5  is  5. 

Proceeding  as  in  the  examples  given  in  §  27,  verify  the 
following:  (+7)  +  (+5)  =  +i2. 

(+7)  +  (-5)=+2. 
(-7)  +  (  +  5)  =  -2.  % 

(_7)  +  (-5)  =  -12. 


POSITIVE  AND  NEGATIVE  NUMBERS  17 

We  see  that  these  answers  can  be  obtained  by  the  following 
rules  which  we  shall  find  very  useful  in  practice : 

The  sum  of  two  numbers  having  the  same  sign  is  found  by  add- 
ing their  absolute  values  and  writing  their  common  sign  before  the 
result. 

The  sum  of  two  numbers  having  opposite  signs  is  found  by  sub- 
tracting the  less  absolute  value  from  the  greater  and  turiting  before 
the  result  the  sign  of  the  number  having  the  greater  absolute  value. 

EXERCISES 

30.  Work  exercises  1-9  by  these  rules  and  then  verify  the 
answers  by  using  the  straight  line  as  in  the  previous  exercises. 

•       1.  (+3)  +  (-6).       4.  (_7)  +  (+17).        7.  (+5*)  +  (4& 

2.  (_3)  +  (+6).       5.  (_17)  +  (-15).      8.  (+5|)  +  (-4> 

3.  (_3)  +  (-6).      6.  (  +  14)  +  (-10).       9.  (-12)  +  (2£). 

10.  Explain  each  of  the  above  answers  when  the  positive 
and  negative  numbers  represent  assets  and  debts;  also  when 
they  represent  temperatures  above  and  below  zero. 

11.  (+6)  +  (+10)+(+6)  +  (-9)=? 

12.  (+50)  +  (-30)4-(+40)  +  (-10)=? 

13.  (_9)  +  (+20)  +  (-3)-f(+20)=? 

14.  (+100)+(+900)  +  (-500)=? 

SUBTRACTION 

31.  If  the  temperature  is  + 10°  in  the  morning  and  +  40° 
at  noon,  we  can  find  the  rise  in  temperature  by  subtracting 
+  10°  from  +  40°.  The  rise  is  (+  40°)  -  ( + 10°)  or  +  30°.  If 
the  temperature  in  the  morning  is  0°,  the  rise  is  (4-40°)  —  (0°) 
or  -f  40°.  If  the  morning  temperature  is  —  5°,  then  the  rise 
must  be  still  greater ;  namely,  +  45°. 

We  have  then  (+  40°)  -(-5°)=  +  45°. 

An  easy  way, to  carry  out  this  subtraction  is  to  change  the 
sign  of  —  5°  to  +5  and  then  to  add  +  5  to  40°. 
c 


18  ELEMENTARY  ALGEBRA 

We  assume  the  following  general  rule : 
To  subtract  a  number,  change  its  sign  and  add. 
This  rule  carries  the  operation  of  subtraction  back  to  that  of  addition. 
Notice  that  we  have  made  no  pretense  of  actually  proving  this  rule  to  be 
true  in  all  cases.     We  take  for  granted  that  it  is.     In  fact,  assuming  this 
rule  amounts  to  a  definition  of  subtraction  in  algebra.     To  the  question, 
what  is  meant  by  "subtracting  a  number,"  the  rule  gives  the  answer : 
"  To  subtract  a  number  is  to  change  its  sign  and  add." 
In  subtracting  a  positive  number  from  a  larger  positive  number,  as 
7  from  12,  it  is  easier  to  follow  at  once  the  familiar  process  used  in  arith- 
metic.    But  the  algebraic  rule  just  given  can   be  applied  to  this  case 
also.     For  we  have 

(  +  12)  -  (  +  7)  =  (  +  12)  +  (-  7)  =  +  5. 

ORAL   EXERCISES 

32.    1.  Verify  the  following  subtractions  : 

+    7         +  22         -7         -7  -20 

-10         +    8         -10         +    8  -    5 

+  17         +14         +   3         -15  -15 

2.  Check  each  answer  in  Ex.  1,  by  adding  the  remainder 
to  the  subtrahend.  What  should  the  sum  be  equal  to  in  each 
case  ? 

3.  Explain  by  the  thermometer  (or  by  assets  and  debts,  or 
by  a  straight  line  with  a  starting  point  0)  how  it  is  possible  to 
subtract  5  from  a  smaller  number  3.  Does  the  introduction  of 
negative  numbers  make  subtraction  always  possible  in  algebra  ? 

Perform  the  following  subtractions : 

4.  +20         5.    +15         6.    -25         7.    -60         8.    +30 
+  30  -30  -40  +10  -50 

9.  8 -(+12).  13.  0-(+10). 

10.  5 -(-40).  14.  0-(-10). 

11.  15- (+60).  15.  (-20) -(  +  20). 

12.  (-30)  -(-10).  16.  50 -(-30). 


POSITIVE   AND  NEGATIVE  NUMBERS  19 

17.  100-  [+30  +  10].  19.    0-(+55  +  15). 

18.  (-5)-  {20  +  40}.  20.    (+20+30) -(+45+25). 

Simplify  the  following: 

21.  5  +  (+ 10)  -  (+ 60).  24.  (_30)-(+10)-(+40). 

22.  6-[  +  9]-[-8]-[  +  7].  25.  85  +  {-90}-{  +  90}. 

23.  (-30)-(-10)-(-40).  26.  85- {  +  90} +  {-90}. 

MULTIPLICATION  INVOLVING  NEGATIVE  NUMBERS 

33.  In  arithmetic  we  define  3  X  4  as  meaning  4  +  4  +  4 ; 
that  is,  4  is  to  be  taken  as  many  times  as  there  are  units  in  3. 
In  the  multiplication  of  fractions,  say,  |x|,  the  above  defini- 
tion becomes  inapplicable,  for  the  reason  that  we  cannot  take 
%  a  fractional  number  of  times.  We  are  therefore  driven  to  a 
different  definition  of  multiplication  ;  we  define  the  product  as 
the  result  obtained  by  multiplying  the  numerators  together 
for  a  new  numerator   and  multiplying  the  denominators  to- 

1     2     1x2 

gether  for  a  new  denominator.     Thus,  oX  £=?, ?* 

A     o     Z  X  O 

In  algebra  we  are  now  studying  a  new  type  of  numbers ; 

namely,  negative  numbers.     When  we  multiply  one  number  by 

another,  and  one  or  both  numbers  are  negative,  the  question 

arises  :  is  the  product  a  positive  number  or  a  negative  number  ? 

We  give  the  answer  to  the  question  in  the  following  definition  : 

The  product  of  two  numbers  having  like  signs  is  a  positive 
number  and  the  product  of  two  numbers  having  unlike  signs  is  a 
negative  number. 

In  multiplying  together  two  numbers,  first  find  the  product 
of  their  absolute  values  and  then  write  before  it  the  proper 
sign. 

Thus  we  have,  according  to  this  definition, 

(  +  3)  x  (+  4)  =  +  12.  (+  3)  x  (-  4)  =  -  12. 

(-3)  x(+4)=  -12.  (-3)x(-4)=+12. 


20 


ELEMENTARY  ALGEBRA 


Illustration.    If  a  man  owes  $  10  to  each  of  two  persons,  he  owes  them 
$  20  all  together.     Denote  the  $  10  owed  by  -  $  10. 
Then  evidently,  (+  2)(-  $ 10)  =  -  $20. 
Second  Illustration.    Denote  a  profit  of  $  10  by  +  $  10. 
Denote  a  loss  of  $  10  by  -  $  10. 
Denote  receive  (add)  by  +. 
Denote  cancel  (subtract)  by  — .     Then 

a.  To  receive  5  profits  of  1 10  each  is  to  be  worth  $60  more.     That  is, 
(+5)(+fl0)=  +$50. 

b.  To  receive  5  losses  of  $  10  each  is  to  be  worth  $  50  less.    That  is, 
(+5)(-$10)=-$50. 

c.  To  cancel  5  profits  of  $  10  each  is  to  be  worth  $  50  Zess.     That  is, 
(-5X+H0)  =  -$50. 

d.  To  cancel  5  Zosses  of  $  10  each  is  to  be  worth  $50  more.     That  is, 
(_5)(-$10)  =  +$50. 


ORAL  EXERCISES 

34.    State  the  following  products  : 


1.    +    5 

2.    -12                3.    -70                4.    +33 

-11 

-12                      +5                      -11 

5.   (+5)x(-5). 

8.    fxf                    11.  [+2|]x[-2^]. 

6.    (_5)x(+5). 

9-    (-l)x(+i).     12.   |-1.2H-1.2}. 

7.   0  x  (-  5). 

10.    (_.5)x(  +  .61).   13.   (+5)(-1.6). 

PROBLEMS 

35.  1.  A  man  saved  $175  a  month  for  4  months;  then, 
during  a  sickness  of  3  months,  he  lost  $200  each  month. 
How  much  did  he  have  at  the  end  of  the  7  months  ? 

2.  During  12  months  a  merchant  expended  $200  each 
month,  and  took  in  $215  monthly.  What  was  his  profit 
during  the  year? 

3.  A  boy  received  from  his  father  on  July  4,  $1.25;  he 
earned  one  dime,  spent  85^  for  firecrackers,  25^  for  a  lunch, 
and  paid  four  5^  car  fares.  How  much  did  he  have  at  the  end 
of  the  day  ? 


POSITIVE  AND  NEGATIVE  NUMBERS  21 

4.  A  road  from  town  A  to  town  B  leads  over  rolling 
country;  5  times  the  road  passes  over  hillocks  rising  each 
37  ft.  and  6  times  it  descends  from  the  hillocks  34  ft.  How 
much  higher  or  lower  is  town  B  than  town  A? 

5.  The  water  in  a  reservoir  rises  3  in.,  then  falls  6  in., 
rises  again  7  in.,  falls  9  in.,  and  finally  rises  13  in.  How 
much  higher  or  lower  is  it  than  at  first  ? 

6.  An  explorer  200  mi.  south  of  the  north  pole  travels  north 
47  mi.,  then  31  mi.  south,  and  again  10  mi.  farther  south,  then 
86  mi.  north.  How  far  north  from  the  starting  point  is  he 
finally  ? 

7.  The  boiling  point  of  water  is  +  212°.  How  much 
above  or  below  the  boiling  point  are  the  following  tempera- 
tures :    +  250°,  +  200°,  +  100°,  +  50°,  -  10°  ? 

8.  Colorado  Springs  is  6000  ft.  above  sea  level.  How 
much  higher  or  lower  are  the  following  elevations  above  sea 
level :  14,000  ft.,  5000  ft.,  400  ft.,  100  ft.,  -  100  ft.  ?  How 
do  you  interpret  —  100  ft.  ? 

9.  The  midday  temperatures  for  one  week  were  4-  39°, 
+  48°,  +  60°,  +  57°,  +  39°,  +  55°,  +  60°.  Find  the  average 
of  these  temperatures. 

The  average  of  a  series  of  numbers  is  found  by  dividing  their  sum  by 
the  number  of  them.  In  this  case  there  are  7  numbers.  Hence  you 
find  their  average  by  dividing  their  sum  by  7. 

10.  The  midnight  temperatures  for  that  same  week  were  as 
follows  :  +  10°,  -  3°,  -  12°,  +  8°,  +  12°,  -  20°,  -  5°.  Find 
the  average  of  these  temperatures. 

11.  A  merchant's  monthly  profits  for  five  consecutive 
months  were  +$400,  +$200,  -$100,  -$300,  +$500. 
Find  the  average  monthly  profits. 

12.  The  latitude  of  Key  West,  Fla.,  is  24°  33'  and  of  Chicago 
is  41°  50'.  What  is  the  latitude  of  a  place  halfway  between 
the  two  ? 


22  ELEMENTARY  ALGEBRA 

13.  The  latitude  of  the  Cape  of  Good  Hope  is  —  34°  21' 
(the  —  indicating  here  south  latitude) ;  the  latitude  of  Athens 
in  Greece  is  37°  58'.  Find  the  latitude  of  a  place  halfway 
between  them. 

14.  At  the  seashore  the  rise  and  fall  of  the  tides  are 
measured  from  a  certain  arbitrarily  chosen  level.  A  tide 
which  falls  below  that  level  is  called  a  "  minus  "  tide.  If  one 
day  the  tide  rises  to  6'  3"  and  then  falls  to  —  V  2",  what  is 
the  average  water  level  for  that  day  ? 

15.  The  Greek  philosopher,  Plato,  died  —  347.  How  many 
years  ago  was  that  ? 

16.  The  elevation  above  sea  level  of  the  top  of  Mont  Blanc 
is  16,050',  and  of  the  lowest  part  of  the  Atlantic  Ocean  is 
—27,800'.    What  is  the  difference  in  elevation  between  the  two? 

17.  A  man  is  rowing  upstream.  In  still  water  his  rate  of 
rowing  is  6  mi.  an  hour.  The  rate  of  the  stream  is  2  mi.  an 
hour.  How  long  will  it  take  him  to  reach  a  place  12  mi.  up- 
stream ? 

18.  If  a  boat  is  steaming  southward  on  a  river  at  the  rate 
of  13  mi.  an  hour,  while  a  man  on  the  deck  is  walking  toward 
the  stern  at  the  rate  of  3  mi.  an  hour,  what  is  the  man's 
actual  motion  with  respect  to  the  shore  ? 

19.  A  balloon  capable  of  exerting  an  upward  pull  of  395  lb. 
is  attached  to  a  car  weighing  146  lb.  What  is  the  net  upward 
or  downward  pull  ? 

DIVISION 

36.  From  the  rule  of  signs  for  multiplication  we  can  ascer- 
tain what  the  rule  of  signs  should  be  for  division. 

In  arithmetic,  we  test  the  correctness  of  a  division  by  mul- 
tiplying the  quotient  by  the  divisor ;  their  product  should  be 
the  dividend.     That  is, 

Divisor  x  Quotient  =  Dividend. 
We  see  that  15  -s-  3  =  5,  because  3  x  5  =  15. 


POSITIVE  AND  NEGATIVE  NUMBERS  23 

Similarly  with  negative  numbers : 
(+ 15)  -=-  (-  3)  =  -  5,  because  (-  3)  x  (-  5)  =  +  15. 
(-  15)  -=-  (  +  3)  =  -  5,  because  (+  3)  x  (-  5)  =  -  15. 
(- 15)  -=-  (  -3)  =  +  5,  because  (-  3)  x  (+  5)  =  -  15. 
From  these  examples  we  see  that  the  rule  of  signs  for  divi- 
sion is  the  same  as  for  multiplication  ;  namely, 

The  quotient  of  two  numbers  having  like  signs  is  a  positive 
number. 

The  quotient  of  two  numbers  having  unlike  signs  is  a  negative 
number. 

From  the  above  it  appears  that  division  is  the  reverse  of 
multiplication.  In  multiplication,  we  are  given  two  factors,  to 
find  the  product ;  in  division,  we  are  given  the  product  and  one 
of  the  factors,  to  find  the  other  factor. 

ORAL   EXERCISES 

37.  Carry  out  the  indicated  divisions  and  check  the 
answers  : 

1.  (+36)  +  (  +  12).   4.  (-42) -=-(-7).    7.  (-51) +  (+17). 

2.  (_64)-=-(+16).   5.  (7.5) +  (1.5).  8.  (+5.1)  +  (-  1.7). 
3-  (-f)-K-f).      6.  (-1.44)+(+1.2).  9.  (+|)^(+.5). 

Perform  the  indicated  operations : 

10.  (+45)  +(-9)  -(-6). 

11.  (+10) -(-7) -(+3). 

12.  (_4) -(-2) -(+5) +(+3). 

13.  (+9)-i-(-3)x(+12)x(-l). 

14.  (-18).  (+3)  +  (-6)  -(-4). 

15.  (+50)x(-i)^(-5  +  7). 

16.  (-2x)-(+3x)  +  (-6x). 

17.  (-8).  (-5)  +  (-10). 

18.  (_i6)-i-(+2)x(-9)+(-8). 

19.  (_24)  +  (-24)x(-24)-(+24). 


24  ELEMENTARY  ALGEBRA 

PROBLEMS  . 

38.  Use  positive  and  negative  numbers  in  the  solutions  of 
the  following  problems : 

1.  A  man  has  a  contract  to  dig  a  well  56  ft.  deep.  If  he 
digs  on  an  average  4  ft.  a  day,  how  many  days  will  it  take  him 
to  dig  the  well  ? 

2.  If  the  foundations  of  a  wall  are  5  ft.  below  the  surface 
and  the  wall  is  41  ft.  high  above  the  surface,  how  many  feet 
are  there  from  the  foundations  to  the  top  of  the  wall  ?  How 
many  cubic  feet  of  brick  in  it  if  it  is  20  ft.  long  and  2  ft. 
thick  ? 

A  SIMPLIFIED  NOTATION 

39.  We  have  seen  that  the  expression 

(+  8)  +  (-  5) 

means  "to  positive  8  add  negative  5."     The  +  between  the 
two  parentheses  means  addition;  the  +  and  —  in  (+8)  and 
(—5)  serve  the  purpose  of  showing  the  quality  of  the  num- 
bers; namely,  that  the  8  is  positive  and  the  5  is  negative. 
We  know  by  our  rules  for  addition  and  subtraction  that 

(+8) +(-5)  =  8-5  =  3  and 

(+8) -(+5)  =  8 -5  =  3. 

This  shows  that  we  can  simplify  the  above  notation  by 
writing,  8  _  5 

and  that  8  —  5  may  have  two  interpretations.     It  may  signify 
either  (+8)_(+5)     or     (+8)  +  (_5). 

In  the  first  interpretation  the  —  sign  means  an  operation; 
in  the  second  interpretation  the  —  sign  means  a  quality. 
The  same  argument  shows  that 

(+3)  -(-4)  =  3  +  4  =  7  and 

(+3)  +  (+4)  =  3  +  4  =  7. 


POSITIVE  AND  NEGATIVE  NUMBERS  25 

We  can  therefore  simplify  expressions  as  in  the  following 
examples : 

(+9) -(-2) +  (+6) -(+7)  =9 +  2+ 6- 7. 
(-6)  +  (+9)-  (-8)  -(+3)  =-6  +  9  +  8-  3. 
The  sign  +  may  be  omitted  in  expressions  involving  multi- 
plications and.  divisions  like  the  following : 

For  (+  27)  •  (+  9)  we  may  write  27  ■  9. 

For  (-  27)  -r-  (+9)  we  may  write  (-  27)  ■+■  9. 

For  (+  15)  -t-  (-f  5)  we  may  write  15  -=-  5. 

EXERCISES 

40.  Simplify  the  following  expressions  : 

1.  (-5) +  (+9) -(-12) -(5).        , 

2.  (+4)(+8)  +  (-6)-(+12). 

3.  (_  24)  -  (+  30)  +  (+  36)  ^  (+  7). 

4.  (-35).  (10) -(+8) -(-7). 

5.  (+18) -(+6) -(+5) +  (+5). 

6.  (24  +  12)  -(-42)  +  (  +  30)  -(-28). 

7.  (4  +  3-2)- (1  +  6) +  (-1  +  8) -[-8-2  + 3]. 

8.  3(3+l)-4(5+8)  +  (2-7)-(5  +  6-7)  +9(8-5). 

9.  10J10-6  +  1J  +  [7  +  3-12]  .4  +  5-15+(3  +  4)5. 

SIMILAR  TERMS 

41.  Terms  which  have  the  same  literal  factors  are  called 
similar. 

Thus,  12  a  and  —  5a  are  similar;  so  are  —  13  xy2  and  25 xy2. 

On  the  other  hand,  terms  which  do  not  coDtain  the  same  literal 
factors  are  called  dissimilar. 

12  a  and  15  b  are  dissimilar  terms;  so  are  —  lSx2y  and  23 xy2. 

If /signifies  "feet,"  we  know  from  arithmetic  that 
10/+ 5/=  15/. 


26  ELEMENTARY  ALGEBRA 

We  find  the  sum  by  adding  the  numerical  coefficients.  This  mode 
of  procedure  is  general.  If /means  "forks"  or  if  /  stands  for 
any  abstract  number,  as  12,  the  same  process  of  addition  holds. 
On  the  other  hand,  if  the  terms  are  dissimilar,  as  5/  and 
10  i,  then  their  sum  is  not  15/  nor  is  it  15  i ;  all  we  can  do  in 
such  a  case  is  to  indicate  the  addition  by  writing 

10/+  5  i.  . 

If  one  boy  has  5  ducks  +  6  hens  +  3  rabbits,  and  another  boy 
has  3  ducks  4-  7  hens  +  8  rabbits,  then  the  two  together  have 
8  ducks  +  13  hens  +  11  rabbits.     Abbreviating,  we  write 

(5  d  +  6  h  +  3  r)  +  (3  d  +  7  h  +  8  r)  =  8  d  +  13  h  +  11  r. 

This  result  is  true,  no  matter  what  d,  h,  r  may  mean. 

Similar  remarks  apply  to  the  subtraction  of  similar  and  dis- 
similar terms. 

EXERCISES 

42.  Find  the  sums  in  exercises  1-11 : 

1.  3a  +  4a  +  7a  —  5a  +  6a. 

2.  (+  5c)  -(+3c)  -5c. 

3.  3x  +  3y  +  4z  +  9y  +  z  +  2x. 

4.  —  3x  +  6x  +  8x  —  6x. 

5.  (+7x)  +  (3x  +  4:y)  +  6y  +  4:X. 

6.  (20c-10d  +  30e)+(4d-5c  +  9e). 

7.  (3  a  +  5  b  -  6  c)  +  (10  b  +  4  e  -  2  a)  +  6  c  -  9  a  +  20  6. 

8.  5  (a  +  6)  +  3  (a  +  6)  +  7  (a  +  6)  -  (a  +  6). 
9.-5  a262  +  4  a&2  -  2  a262  -  6  a&2  +  7  a62c  +  0  a6»c. 

10-    -r-  3/flr,  +  4  grA,  -  5  <9/t,  -  7  grft,  +  6/7. 

11.  2(a  +  &)  +  c,  3  a,  76,  9  c. 

12.  From  —  8  ax  take  —  5  cue,  then  add  to  the  result  —  6  ax. 


POSITIVE  AND  NEGATIVE  NUMBERS  27 

13.  Add  9  inn  to  +  4wm;  from  the  result  take  10  ran. 

14.  Subtract  +  6  kl  from  —  24  kl,  then  add  +  12  kl. 

15.  Subtract  3  be  from  —  5  ac,  then  add  —  9  cd. 

16.  Simplify  :  40  st  -  5  st  +  20  st  +  ty  -  10  ty. 

17.  From  6  (a  4-  x)  take  7  #,  then  again  —  4  a. 

ADDITION   AND  SUBTRACTION  OF  POLYNOMIALS 

43.  When  polynomials  are  to  be  added  together,  or  when 
one  polynomial  is  to  be  subtracted  from  another,  it  is  conven- 
ient to  write  similar  terms  in  the  same  column  and  to  add  or 
subtract  the  terms  in  each  column,  proceeding  from  left  to 
right. 

Add  2o  +  46-5  c,  -2a  +  46  +  7c,  —  8a-56  +  9c. 
Solution.     2  a  +  ib  —    5c 
-2a  +  4b  +    7  c 

-  8a-  55  +    8c 

—  8a  +  36  +  10c,  the  required  sum. 
From  9 x  +  8y  —  hz  take  6x  —  7  y  +  \z. 

Solution.  9a;-f  8y—  hz 
6x  —  ly  +  4s 
3  x  +  15  y  —  9  z,  the  required  difference. 

EXERCISES 

44.  Add  the  following  polynomials : 

1.  2  ac2  +  3  h2c  -  c3,  6ac  -  b2c  +  9  ac2,  ac  +  b2c-5  c3. 

2.  a2  +  8  b2  +  4  c2,  2  a2  -  5  62  -  8  c2,  662  +  7c2. 

3.  6  x  —  7  y  +  4  a#,  —  4  a  -+  5  y  +  10  xy,  20  a  —  9  xy  +  y. 

4.  3a:  +  ?/-8z  +  3w,  9a;  —  8  ?/  —  6  z  +  4  to  —  3  v. 

5.  a  +  &,  a  +  c,  6  +  c,  2  a  —  3  b  +  4  c. 

6.  a  +  b  +  c  —  d,  2  b  +  3 c  +  7  d,  ±a  +  5  b  -  6  d. 

7.  la+6  6  +  3c-10d+/,  a-5  6+8  c+2/,  2  6+.3c+4d. 

8.  a3b  +  2ab3-  6ab,  3a3b-oab  +  9  ab3. 


28  ELEMENTARY  ALGEBRA 

In  exercises  9-11,  subtract  the  second  polynomial  from  the 
first  and  check  your  answer  by  adding  the  remainder  to  the 
subtrahend : 

9.    a  +  6  +  c  +  d  +  e,  2a  +  36-4c  —  3  e. 

10.  10  x  +  10  y  —  10  z  +  10  w,  5  w  +  3  x  —  8  y  +  3  z  —  v. 

11.  a +  6  —  x-\-y  +  3z,  a  +  b  +x  —  y  —  Sz  +  5x. 

12.  Toa  +  c-d  add  4a  +  5c  +  6a*  and  from  the  sum  take 
3cH  6c  +  9d. 

13.  From  2m+4n-f  9/)  take  9  m  —  Sn  +  T p  and  to  the  re- 
sult add  4  ra  +  4  «  +  4|>. 

14.  From  27/  +  24  a  +  23  ft  take  19/- 15  h  + 15  gr  and  from 
the  result  take  20/+  3  g  —  9  h. 

15.  Subtract  26  +  4c+7d  +  6e  from  56  —  4a*  +  6c  +  9e 
and  check. 

16.  From  3  a  +  r  +  s  take  the  sum  of  10  r  +  30  s  and  5  r  — 
7  s  +  20  a. 

17.  From  the  sum  of  a  +  3  b  and  2  a  —  6  6  take  the  sum  of 
—  3  a  +  24  b  and  a  +  b  +  c. 

18.  From  a;  +  y  +  z  +  w>  take  a  —  b  —  c  —  d,  then  add  to  the 
result  3z  +  4w-7c+3c7. 

19.  From  4  (a  +  ft)  take  6  (a  +  b  +  5  c). 

20.  Add  2  (a  +  4  b  +  3  c),  3  (-  2  a  +  3  6  +  7  c),  4  (6  +  c). 

21.  Add  4  a4  +  4  a  +  a46,  0  a4  —  3  a  +  5  a46,  —  6  a*b,  -9  a. 

Evaluate  the  following  monomials  and  polynomials,  when 
a  =  -  2,  b  =  -  3,  c  =  - 1,  d  =  10 : 

22.  a6,  a6c,  be,  2ab,  —  2  a6,  —  3  be,  3  6c,  ctf". 

23.  a  +  6+c  +  d,  a  —  6  +  c  +  2  a",  —  5a  —  36  +  ac. 

24.  a2,  62,  c2,  cP,  a262. 

25.  abed,  a2bcd,  ab2cd,  ab<?d,  abed?. 

26.  a3,  63,  c3,  d3,  a363,  a3**3,  c3^. 

27.  aV,  a?d?,  bd3,  cd?,  <?d?. 


POSITIVE  AND  NEGATIVE  NUMBERS  29 

CHECKING  BY  SUBSTITUTION  OF  NUMBERS 
EXERCISES 

45.  1.    Add  3 x2- 4 y,  -x2  +  7y,  5x2-12Sy. 

In  the  polynomials  added  and  in  their  3  x2  —     4  y  =      3—      8 

sum,  let  each  letter  equal  some  simple  —  x2  -f      7y  =  —  1  +    14 

number  and  compare  as  indicated  below.  5x2  —  128  y  _      5  —  256 

If  a  mistake  has  been  made  in  addition,       Sum  7  x2  —  125  y  7  —  250 

this  test  almost  always  reveals  it.  7  x2  —  125  y  =  —  243. 

In  each  of  the  three  binomials  and 
also  in  their  sum,  let  x  =  1,  y  =  2.     We  obtain  the  numbers  on  the  right. 
Adding  the  two  columns  of  numbers,  the  sums  are  7  —  250.     This  is  the 
very  same  result  as  that  obtained  by  letting  x  =  1  and  y  =  2  in  the  sum 
7  x2  —  125  y.     This  checks  the  addition. 

Add  the  following  and  check  by  letting  x  =  2,  y  =  l: 

2.  5a;2  —  6x  +  4y,  x2  +  7x  —  3y,  —  3x2  —  5x  —  2y. 

3.  -x2  —  7x-2y,  I0x2  +  8x  —  4y,  —  6x2  +  ±x  -  lOy. 

4.  10  x2  —  x  -  y,  11  x2  -  8  x  —  9  y,  —  20  x2  +  7  x  -  5  y. 

In  exercises  5-7,  subtract  the  second  polynomial  from  the 
first  and  check  by  letting  a  =  2,  b  =  3 : 

5.  10a2  +  a  —  b,  5a2  —  4a  +  6. 

6.  25  a2-  10 a  +  2  6,  20a2 -11  a  +  3  6. 

7.  13a2  +  14 a  +  15 6,  20 a2 -15 a  +  6. 

8.  10  a3 +  2  ft2,  7a3-3  62. 

9.  12  a3  -  5  63,  7  a3  -  7  63. 
10.    7  a3  -  24  b\  8  a3  -  20  64. 

REMOVAL  OF  PARENTHESES 

46.  In  an  example  like  2  a  +(a  +  b  +  c  —  d),  the  polynomial 
within  the  parenthesis  is  to  be  added  to  what  precedes.  This 
may  be  accomplished  by  adding  each  term  separately.  We 
may  at  first  merely  indicate  this  separate  addition  by  rewriting 
the  expression  with  the  parenthesis  omitted ;  thus, , 

2a+(a  +  6  +  c-d)=2a  +  a  +  6  +  c-cZ. 


30  ELEMENTARY  ALGEBRA 

Thereupon   like   terms   are   combined.     In   this   instance    we 
obtain,  3  a  +  b  +  c  —  d. 

In  an  example  like  2  a  —  (a +  6  +  c  —  d),  the  polynomial 
within  the  parenthesis  is  to  be  subtracted  from  what  precedes. 
This  may  be  accomplished  by  subtracting  each  term  separately. 
We  may  at  first  merely  indicate  this  subtraction  by  rewriting 
the  expression,  but  changing  the  sign  of  each  term  as  it  is 
taken  out  of  the  parenthesis ;  thus, 

2a  —  (a  +  b  +  c  —  d)=2a  —  a  —  b  —  c  +  d. 
Thereupon   like   terms   are   combined.     In   this   instance   we 
obtain,  a  —  b  —c  +  d. 

We  have  then  the  following  rules  for  the  removal  of 
parentheses : 

A  parenthesis  preceded  by  the  +  sign  may  be  removed  without 
changing  the  signs  of  the  terms  that  were  within  the  parenthesis. 

A  parenthesis  preceded  by  the  —  sign  may  be  removed  if  we 
change  the  sign  of  each  term  that  was  within  the  parenthesis. 

When  an  expression  contains  a  parenthesis  within  another 
parenthesis,  as  —  (5  a  —  [4  b  —  5  c]),  remove  one  parenthesis 
first,  then  the  other.     Thus, 

-(5  a  -  [4  6  -  5  c])=  -  (5  a  -  4  6  +  5  c)=  -  5  a  +  4  6  -  5  c. 

It  is  usually  found  easier  to  remove  the  inner  parentheses 
first. 

The  student  should  form  the  habit  of  guarding  carefully 
against  error  in  the  application  of  these  simple  rules. 

EXERCISES 

47.    Kemove  the  parentheses  and  simplify  : 

1.  a  +(3  a  +  5  —  4  c). 

2.  b  -(6  +  2c  +  3d). 

3.  (a  +  b  +  3c)  +  (a  +  b)-(a-b). 

4.  (a  +  4&-5c)-(3a  +  46  +  2c)  +  (a  +  &). 

5.  [10.<z-2n  +  4g]-£10a;  +  2n-f 4?}. 


POSITIVE   AND   NEGATIVE  NUMBERS  31 

6.  -(_3d  +  4e-5/)-(3e  +  7/-6<7). 

7.  {_91+8m— 7w}+[— 71+6n+4m]  — (3r  +  5*-9y). 

8.  [2x  +  (2y  +  z)  +  w]-\5y-(2x-z)\+(10x  +  9y). 

9.  [5a  +  6|-{2a-36|-9a+6  6]-(4a4-56). 

10.  -(-p  +  g)-(4g  +  6p  +  3r)-(-5p  +  r-5g). 

11.  6c-(4d  +  [e-3/]). 

12.  4a  +  {-26-(c-4d)+«|. 

13.  (_2a  +  |46-5c|-|26-3c|  +  6a). 

14.  5  6-(4c-a)  +  (-3  6  +  c)-(5a-5&  +  c|  +  3c). 

15.  —  (m  —  n)  +  (3  m  +  4  n)  —  {  —  m  +  (3  m  —  n)  J . 

16.  \  —  (ra  -f  4  n  —  p) —  (p  —  2  n)  +  (4  p  —  3m)—  7  m  j . 

INSERTION   OF  PARENTHESES 

48.  This  process  is  the  inverse  of  the  preceding.  Since  in 
removing  a  parenthesis  with  the  +  sign  before  it  the  signs  of 
the  terms  taken  out  are  not  changed,  it  follows  that  when  the 
parenthesis  is  restored,  the  signs  of  the  terms  inserted  in 
the  parenthesis  are  not  changed,  provided  the  sign  before 
the  parenthesis  is  +• 

Since  in  removing  a  parenthesis  with  the  —  sign  before  it  the 
sign  of  every  term  taken  out  is  changed,  it  follows  that  when 
the  parenthesis  is  inserted,  the  sign  of  every  term  in  the  pa- 
renthesis must  be  changed,  provided  the  sign  before  the  paren- 
thesis is  — . 

Accordingly      a  +  b  —  c  +  d  =  a  +  (b  —  c  +  d),  and 

*=  a  —  (—  b  +  c  —  d). 

EXERCISES 

49.  Inclose  in  parentheses  all  the  terms  except  the  ones  in- 
volving the  letters  x  and  y ;  in  the  odd-numbered  exercises  use 
the  -f-  sign  before  the  parentheses ;  in  the  even-numbered  exer- 
cises use  the  —  sign. 

1.  x  +  a  +  6—  c  +  d  +  y  —  m  +  n  —  s. 

2.  2x  +  3a  —  ±b  +  5c-y-6k  +  8l  +  7f—  4g. 


32 


ELEMENTARY   ALGEBRA 


-i- 


i  / 
b<?- 


-k- 


Fig.  4 


3.  3  a+2  w—3  e+x— 4  r  +5 1+  y+  5t  —  6  u  +  7  i— x+a+b. 

±.  —9m  +  8i-7u  +  6y  +  5t  —  4:r  +  3e  +  x  +  2w  +  3a. 

5.  3y  —  c  +  3a-2x  +  4:b-2d. 

6.  —  5  &  —  cc  +  3?/  —  4  c  +  5  d. 

7.  —  5m  +  6n  +  7a  —  3p  —  4y. 

8.  3y  —  3c  +  26  —  5a  +  5o;. 

MAGIC  SQUARES 

4.  a  a  ,-*  1.  Figure  4  is  a  magic  square.  Add 
the  numbers  along  each  of  the  three 
-i'-\ — y  horizontal  dotted  lines,  each  of  the 
three  vertical  lines,  and  each  of  the  two 
diagonal  lines.  What  is  there  peculiar 
— >  about  the  sums  ? 
2.  This  magic 
square  (Fig.  5) 
has  algebraic  expressions  arranged 
along  the  eight  lines.  Find  the  sum 
of  the  three  algebraical  expressions 
along  each  of  these  lines.  What  is 
there  peculiar  about  the  sums  ? 

3.   Take  a  =  1,  b  =2,  and  compute 
the  resulting  numerical  magic  square. 

Select  other  integral  values  for  a  and  b,  and  construct  other 

magic  squares. 

4.  If  a  be  2  and  6  =  1,  then 
a  magic  square  is  obtained, 
most  of  whose  numbers  are 
negative.  Is  the  sum  of  the 
numbers  still  the  same  for 
each  of  the  8  different  lines  ? 

5.  Find  the  sum  of  the 
expressions  in  each  of  the  10 
lines    in    the    larger    magic 

Fig.  6  square  (Fig.  6). 


3  6 
-2a 

86 
—  la 

6 

26 
—  a 

46 
-3a 

6  6 
—  5  a 

76 
—  6a 

a 

56 
-4a 

Fig.  5 


a 

14  6 
-13  a 

13  6 
-12  a 

36 

-2a 

116 
-  10  a 

56 
-4a 

66 
—  5  a 

86 

-7a 

76 
-6a 

96 

-8a 

10  6 
-9a 

46 
-3a 

12  6 
-  11a 

26 

—  a 

6 

15  6 
-  14  a 

POSITIVE  AND  NEGATIVE  NUMBERS  33 

6.    Answer  questions  2  and  3,  given  above,  for  this  larger 
magic  square. 

ORAL    EXERCISES 

50.  Represent  a  number : 

1.  Greater  than  n  by  2.  6.    5  times  x  less  2. 

2.  Greater  than  x  by  5.  7.    10  greater  than  a  +  b. 

3.  Less  than  y  by  3.  8.    x  greater  than  9. 

4.  Less  than  r  by  6.  9.  a  greater  than  n  —  x. 

5.  Less  than  7  n  by  a.  10.   y  less  than  2  times  x. 

What  is  the  other  part  if 

11.  One  part  of  15  is  6  ?  14.  One  part  of  y  is  x? 

12.  One  part  of  n  is  2  ?  15.  One  part  of  24  is  a  ? 

13.  One  part  of  x  is  y  ?  16.  One  part  of  a  +  b  is  10  ? 

17.  The  difference  between  two  numbers  is  16;  the  greater 
is  20.     What  is  the  smaller  ? 

18.  The  difference  between  two  numbers  is  a;  the  greater 
is  x.     What  is  the  smaller  ? 

19.  The  smaller  of  two  numbers  is  y ;  their  difference  is 
3.     What  is  the  greater  ? 

20.  The  sum  of  two  numbers  is  27  ;  the  smaller  is  n.     What 
is  the  greater  ? 

21.  The  sum  of  two  numbers  is  a ;  the  greater  is  y.     What 
is  the  smaller  ? 

22.  By  how  much  does  28  exceed  15  ? 

23.  By  how  much  does  x  exceed  11  ? 

24.  Express  the  excess  of  52  over  45. 

25.  Express  the  excess  of  x  oveivy- 

26.  Express  the  excess  of  a  +  b  over  a  —  b. 

27.  If  one  book  costs  $  2,  what  will  7  books  cost  ? 

28.  If  one  pencil  costs  5  ^,  what  will  b  pencils  cost  ? 


34  ELEMENTARY   ALGEBRA 

29.  If  one  coat  cost  x  dollars,  what  will  3  coats  cost  ? 

30.  If  7  hats  cost  $  17i  what  will  1  hat  cost  ? 

31.  If  2  oranges  cost  d  cents,  what  will  1  orange  cost  ? 

32.  If  x  apples  cost  15^,  what  will  1  apple  cost  ? 

33.  What  is  5  %  of  $200  ?     4  %  of  d  dollars  ? 

34.  What  is  the  interest  on  $  250  at  6  %  ? 

35.  What  is  the  interest  on  $  300  at  r  %  ? 

36.  What  is  the  interest  on  p  dollars  at  r  °fo  ? 

37.  If  a  man  is  53  years  old  now,  how  can  you  represent  his 
age  x  years  hence  ? 

38.  If  a  man's  age  now  is  n  years,  how  can  you  represent 
his  age  20  years  ago  ? 

39.  If  a  man  is  twice  as  old  as  his  son,  and  his  son  is  b 
years  old,  how  old  is  the  father  ? 

40.  The  side  of  a  square  is  6  in.     What  is  its  perimeter  ? 

41.  The  side  of  a  square  is  x  in.     What  is  its  perimeter  ? 

42.  The  length  of  a  rectangle  is  twice  its  width.  Express 
its  length  and  its  perimeter,  if  w  represents  its  width. 

43.  The  length  of  a  rectangle  is  4  ft.  more  than  twice  its 
width.  Express  its  length  and  its  perimeter  if  x  represents 
its  width. 

44.  If  I  have  x  dimes,  how  can  I  express  the  number  of  cents 
I  have  ? 

45.  If  n  represents  a  certain  number,  how  can  I  represent 
the  next  greater  number  ? 

46.  If  n  represents  a  certain  even  number,  how  can  I  repre- 
sent the  next  even  number  ? 

47.  Express  two  consecutive  odd  numbers. 


POSITIVE  AND  NEGATIVE  NUMBERS  35 

SOLUTION  OF  EQUATIONS 
51.   We  have  seen  that  an  equation  is  like  a  balance  (§  5)  ; 
the  equality  or  balance  is  not  disturbed  as  long  as  the  same 
changes  are  made  simultaneously  on  both  sides. 

Thus,  in  solving  9x  —  10  =  4  x. 
Adding  10  to  both  sides,  9  x  =  4x  +  10. 

Subtracting  4x  from  both  sides,  5x  =  10. 

Dividing  both  sides  by  5,  x  =  2,  the  answer. 

It  will  be  noticed  that  adding  the  10  to  both  sides  has  the 
same  effect  as  simply  moving  the  10  to  the  other  side  of  the 
equation  and  changing  its  sign. 

In  the  same  way,  subtracting  4  x  from  both  sides  had  the  same 
effect  as  taking  4  x  over  to  the  left-hand  side  and  changing  its 
sign. 

This  operation  is  called  transposition.  It  consists  in  moving 
a  term  from  one  side  of  the  equation  to  the  other  and  changing  the 
sign  of  the  term. 

Thus,  in  solving  4  —  6  x  =  2  x  —  36. 

Transposing  4  yields  —  6  x  =  2  x  —  36  —  4. 

Or  -  6  x  =  2  x  -  40. 

Transposing  2  x  gives  —  8  x  =  —  40. 

Dividing  both  sides  by  —  8,  x  =  5. 

Check  :  Substitute  5  for  x  in  the  original  equation.     The  result  is 
4  -  6  •  5  =  2  •  5  -  36,  or    4  -  30  =  10  -  36,  or 
-  26  =  -26 
Since  this  is  correct,  we  know  that  5  is  the  correct  value  for  x,  provided 
that  no  mistake  was  made  in  checking. 

The  method  of  solution  which  we  have  explained  applies 
equally  well  when  the  coefficients  of  the  equation  are  numbers 
represented  by  letters. 

Consider,  for  example,  mx  +  b  —  c  =  d  —  mx. 

Transpose  —  mx,  +  b,  —  c,  mx  +  mx  —  d  —  b  +  c. 

Combine  mx  +  mx,  2  mx  =  d—  b  +  c. 

Divide  both  sides  of  the  equation  by  2  m,    x  =  — ,  the  answer. 

2m 


36  ELEMENTARY   ALGEBRA 

The  answer  obtained  in  the  solution  of  an  equation  is  called 
a  root  of  that  equation. 

A  root  of  an  equation  is  a  quantity  which,  when  substituted 
for  the  unknown  in  the  equation,  "  satisfies  "  the  equation  by 
reducing  both  members  to  identical  numbers. 

EXERCISES 

52.  Solve  the  following  equations,  and  check  each  answer : 

1.  6x4-5  =  3x4-17.  11.    50#  +  20?/  +  40  = 

2.  13x-5  =  llx4-5.  ±Q-3y  +  70-7y. 

3.  4-5*=  -6x4-2.  12-   Ji  +  10.+  |f-86- 

a      a  on        on        a  #  Z  4-  50  —  45. 

4.  4  x  —  30  =  30  —  6  x.  ,00      ,», 

13.    2  x  +  o  =  a. 

5.  3y  +  S  =  5y-5.  14    2x  +  6  =  a>  +  a -a 

6.  6y  +  ±5  =  7y  +  50.  15    fta,  +  6  =  c  _  ^ 

7.  40  +  4  =  90-6^.  16    5x  =  a4-&-*. 

8.  l+X-6x  +  30=  17.     |2/_a=  -b-ly, 

a  4- 35- 3  x.      is.   aa;  +  4  =  5. 

9.  £a  +  5  =  6.  19.    ex -3c  =  10c. 
10.   £x- 10  =  2x4- 20.            20.    az  +  2  6  =  3  c. 

53.  In  a  balance  the  equilibrium  is  not  disturbed  if  we  sub- 
stitute on  one  side,  say,  a  10-pound  weight  in  place  of  a  4- 
pound  weight  and  a  6-pound  weight.  So,  in  an  equation,  it  is 
permissible  to  leave  one  side  unaltered  and  to  make  any  change 
on  the  other  side  we  wish,  provided  this  change  does  not  alter 
the  value  of  that  other  side. 

Only  in  an  operation  which  changes  the  value  of  one  side  of 
a  balance  or  equation  we  must  be  careful  that  exactly  the  same 
change  in  value  is  made  on  the  other  side.     For  example : 

2  (4  x  +  6)  -  3  x  -  4  -  3  (5  x  -  9)  +  13a;. 
8  x  +  10  -  3  x  =  4  -  15  x  +  27  +  13  x. 
5x  +  10  =  -2a5  +  31. 
7  a;  =  21. 
a;  =  3. 


POSITIVE  AND  NEGATIVE   NUMBERS  37 

EXERCISES 

54.  Solve: 

1.  2(5z  +  6)  +  20  =  4a;  +  5(a;  +  2). 

2.  K6^8)-^*1^  6(z +  4) -21. 

3.  lOz- 20 +  25  =  ^(10z  +  20) -60. 

4.  {(x  +  2)  =  x-20. 

6.   3(aj-5)-2(x  +  2)  =  5(a;-7). 

PROBLEMS 

55.  1.  The  length  of  a  rectangle  exceeds  its  width  by  10  ft. 
Its  perimeter  is  52  ft.     Find  its  length  and  width. 

2.  The  length  of  a  rectangle  is  70  in.  less  than  its  perime- 
ter.    Find  its  length,  when  its  width  is  20  in. 

3.  A  lot  is  3  times  as  long  as  it  is  wide.  If  the  perimeter 
is  480  ft.,  what  are  its  dimensions  ? 

4.  A  garden  in  the  form  of  a  rectangle  is  70  yd.  wide. 
The  perimeter  is  4  times  the  length.     Find  the  length. 

5.  A  grocer  desires  to  mix  two  grades  of  coffee  costing  35^ 
and  45^  a  pound,  so  as  to  obtain  a  mixture  weighing  100  lb. 
and  worth  38^  a  pound.  How  many  pounds  of  each  kind  must 
he  take  ? 

6.  How  many  pounds  of  50^  coffee  must  be  mixed  with 
60  lb.  of  40^  coffee  to  get  a  mixture  worth  43^  a  pound  ? 

7.  Two  grades  of  coffee  costing  35^  and  30^  are  to  be 
mixed  so  as  to  obtain  a  mixture  weighing  100  lb.  and  selling 
for  50  ^  a  pound  at  a  profit  of  50  %  on  the  cost.  How  many 
pounds  of  each  kind  are  needed? 

8.  The  sum  of  two  numbers  is  206 ;  the  larger  exceeds  the 
smaller  by  46.     Find  the  numbers. 

9.  What  two  numbers  have  216  for  their  sum  and  300  for 
*  their  difference  ? 


348554 


38  ELEMENTARY   ALGEBRA 

10.  The  difference  between  two  numbers  is  25 ;  their  sum  is 
0.     Find  the  numbers. 

11.  If  the  difference  between  two  numbers  is  24,  what  must 
each  number  be  in  order  that  their  sum  shall  be  16  ? 

12.  Two  bicyclists  start  in  the  same  direction,  one  going 
at  the  rate  of  10  mi.  an  hour  and  the  other  at  the  rate  of  12  mi. 
an  hour.  How  long  before  the  second  will  be  1  mi.  in  advance 
of  the  first  ? 

Let  x  =  the  no.  of  hr.  required. 
Then  10  x  =  the  no.  of  mi.  the  first  goes  in  x  hr. 

12  x  =  the  no.  of  mi.  the  second  goes  in  x  hr. 
In  x  hr.  the  second  has  gone  1  mi.  farther  than  the  first.     Hence  if 
we  add  1  mi.  to  the  distance  the  first  has  traveled,  we  get  12  x,  the  dis- 
tance traveled  by  the  second.     That  is, 

10  x  +  1  =  12  x. 
Solving  this,  we  get  x  =  §. 

13.  Two  boys  start  to  run  in  the  same  direction.  One  runs 
at  the  rate  of  300  yd.  a  minute ;  the  other  at  a  rate  of  340  yd. 
a  minute.  How  long  before  the  second  will  be  half  a  mile  in 
advance  of  the  first  ? 

14.  If  the  two  boys  in  problem  13  were  to  start  at  the  same 
moment  and  run  in  opposite  directions  on  a  two-mile  circular 
running  track,  in  how  many  minutes  would  they  meet  ? 

15.  A  boy  has  $  1.35  in  his  pocket,  all  in  dimes  and  nickels. 
How  many  coins  of  each  kind  has  he,  if  their  total  number  is 
19? 

16.  How  many  dimes  and  quarters,  16  coins  in  all,  are 
necessary  to  amount  together  to  $  2.65  ? 

17.  A  newsboy  has  collected  $1.99,  the  amount  consisting 
of  nickels  and  pennies.  How  many  has  he  of  each,  the  total 
number  of  coins  being  103  ? 

18.  I  have  $  7.45  in  19  coins,  which  are  quarters,  dimes,  and 
5  one-dollar  pieces.     Tell  the  number  of  coins  of  each  kind. 


POSITIVE  AND   NEGATIVE   NUMBERS  39 

19.  The  average  of  the  highest  and  the  lowest  temperature 
of  a  winter's  day  is  17£° ;  the  difference  between  them  is  55°. 
Find  the  extreme  temperatures. 

20.  If  the  lowest  temperature  one  day  was  —  8°,  and  the 
average  for  that  day  was  20°,  find  the  maximum  temperature. 

21.  A  certain  man  is  worth  $  3000  more  than  when  he 
started  out  in  business ;  one  third  of  the  sum  of  what  he  has 
now  and  what  he  did  have  at  first  is  $  1915.  How  much  does 
he  possess  now  ? 

22.  The  sum  of  two  investments  is  $  2700 ;  the  first  yields 
annually  at  5  °f0  $30  less  than  does  the  second  investment  at 
6  °fo .     Find  each  investment. 

23.  A  man  divides  $5000  into  two  investments,  of  which 
one  at  6  %  brings  annually  $40  more  than  the  other  at  7  %. 
Into  what  sums  was  the  $  5000  divided  ? 

24.  John  can  write  twice  as  fast  on  a  typewriter  as  James 
can  write  by  hand.  How  many  words  can  each  write  per 
minute,  if  together  they  write  162  words  per  minute  ? 

25.  The  sum  of  two  numbers  is  81 ;  their  difference  is  11. 
Find  the  two  numbers. 

26.  The  sum  of  two  numbers  is  4  a  +  b ;  their  difference  is 
a  •+■  b.     Find  the  numbers. 

Let  the  first  number  =  x. 

Then  the  second  number  =  4  a  +  b  —  x. 

The  first  number  subtracted  from  the  second        =  4  a  +  6  —  x  —  x 

=  ia+b-2x. 
By  the  conditions  of  the  problem,  this  difference  =  a  +  b. 
Hence  we  have  the  equation  4  a  +  b  —  2x  =  a  +  b. 

Solve  this  equation. 

27.  The  sum  of  two  numbers  is  a ;  their  difference  is  b.  Find 
the  two  numbers. 

28.  The  difference  between  two  numbers  is  c;  their  sum  is 
0.     Find  the  two  numbers. 


40  ELEMENTARY   ALGEBRA 

29.  How  many  dimes  and  quarters  will  make  a  sum  of 
10  a  +  25  b  cents,  if  the  number  of  coins  is  a  +  b  ? 

30.  If  the  lowest  temperature  one  day  was  —  t  °,  and  the 
average  temperature  was  T°,  what  was  the  maximum  tem- 
perature ? 

31.  A  merchant  owns  now  d  dollars  more  than  when  he 
started  out  in  business ;  one  third  of  the  sum  of  what  he  owns 
now  and  what  he  did  own  at  the  beginning  is  D  dollars.  How 
much  does  he  own  now  ? 

32.  A  father  earns  4  times  as  much  as  his  son.  Together 
they  earn  a  dollars  per  day.     How  much  does  each  earn  ? 

33.  The  perimeter  of  a  rectangle  is  p ;  its  length  exceeds  its 
width  by  q.     Find  the  length  and  breadth. 

34.  A  rectangle  is  3  times  as  long  as  it  is  wide.  Find  its 
dimensions,  if  the  perimeter  is  P. 

35.  Find  two  consecutive  integers  whose  sum  is  25. 

If  the  smaller  integer  is  x,  then  the  next  higher  integer  is  x  +  1 . 

36.  Find  three  consecutive  integers  whose  sum  is  75. 

37.  What  three  consecutive  integers  produce  the  sum  s  ? 

38.  Find  two  consecutive  even  integers  whose  sum  is  102. 

If  x  is  any  integer,  then  we  are  certain  that  2  a;  is  an  even  number,  for 
the  reason  that  any  integer,  when  doubled,  gives  an  even  number. 

Letting  2  x  stand  for  the  smaller  even  integer,  what  must  be  added  to 
this  to  give  the  next  higher  even  integer  ? 


LEONHARD  EULER 


One  of  the  greatest  mathematicians  of  the  eighteenth  century.  In 
1770  he  published  an  algebra  which  was  translated  from  German  into 
French,  English,  Italian,  and  Latin,  and  was  widely  used.  Eider  was 
a  Swiss,  but  spent  most  of  his  time  in  Russia  and  Germany. 


CHAPTER   III 

MULTIPLICATION   AND   DIVISION 

MULTIPLICATION  OF  MONOMIALS 

56.  By  the  definition  of  an  exponent  (§  14),  we  know  that  a2 
means  a -a,  b3  means  b -b -b;  hence 

a2  x  a3  means  a  •  a  x  a  •  a  •  a  or  a5.     Similarly, 

c3  x  c*  =  c7,  and  c?V&  x  a2b3c5  =  abb5c9. 

Hence  in  the  multiplication  of  terms  containing  like  letters,  the 
exponent  of  any  letter  in  the  product  is  the  sum  of  the  exponents 
of  that  letter  in  the  factors. 

The  product  of  12  abc  and  5  ab2c  is  60  a2b3(? ;  that  is,  the 
coefficients  are  multiplied  together.  Bearing  in  mind  the  rule 
of  signs  in  multiplication,  the  following  results  are  seen  to  hold : 
+  3  xy  x  ( +  5  xhjz)  =  -f  15  x*y2z. 

—  4 xy2  x  ( +  6  x2yz2)  =—2A  x^ffz2. 
+  5  xyz2  x  (—  6  x?y2z)  =  —  30  tfifz3. 

—  6  xyz  x  (—  7  xyz)  =4-42  artyV. 

ORAL   EXERCISES 

57.  Perform  the  following  indicated  multiplications : 

1.  +3x(  +  5xy).  7.  +  50  df3  x  -  9f2gh. 

2.  +5xz-(-5  xy*z).  8.  +  20  times  +  30  e\ 

3.  -7  a3b2c  times-  8  6W.  9.  -  30  acf  x  - 10  aWcdf3. 

4.  +  a>Y2*  times  60.  10.  (-2xy)-(+4:x2y)-(-lxyz). 

5.  (8  mln*o3)  ■  (-  7  wiVo2).  11.  (-  2  a)  (-  3  a2)  (-  2  ab). 

6.  -  12  /i8  x  -  7  h*jk.  12.  (+2  aft)  (-2  a2b)  (-  3  aft3). 

13.  If  two,  four,  or  six  negative  terms  are  multiplied  to- 
gether, what  is  the  sign  of  the  product  ?     Why  ? 

41 


42  ELEMENTARY   ALGEBRA 

14.  If  three,  five,  or  seven  negative  terms  are  multiplied 
together,  what  is  the  sign  of  the  product  ?     Why  ? 

15.  If  among  the  terms  to  be  multiplied  together  a  certain 
number  of  terms  is  positive  and  the  number  of  negative  terms 
is  even,  what  is  the  sign  of  the  product  ?  Does  the  number  of 
positive  terms  require  special  attention  ? 

16.  If  the  number  of  negative  terms  is  odd,  what  is  the  sign 
of  the  product  ? 

MULTIPLICATION  OF  A  POLYNOMIAL  BY  A  MONOMIAL 

58.  It  is  seen  that         4(2  +  3)  =4  -5  =  20, 
and  that  4  •  2  +  4  •  3  =  20. 

Hence  4(2  +  3)=4 -2 +  4 -3. 

In  general,  a(b  +  c)  =  ab  +  ac, 

and  a(b  +  c  —  d)=ab  +  ac  —  ad. 

Hence,  in  multiplication  of  a  polynomial  by  a  monomial,  we 
have  the  rule : 

Multiply  each  term  of  the  polynomial  by  the  monomial,  and 
write  down  in  succession  the  products  with  their  respective  signs. 

Observe  that  the»  foil  owing  results  hold: 

+2  ab\a2  +  b  -  c3)  =  +  2  a362  +  2  ab3  -  2  at**. 
-  3  a?c[  -  a  +  2  be  -  4  c]  =  +  3  a3c  -  6  atbc2  + 12  a2^ . 

EXERCISES 

59.  1.    Multiply  2  x  +  ey  +  6  z  —  5  w  by  3  xy. 

2.  Multiply  a  —  26  +  ec  —  4d  —  ebylO abc. 

3.  Find  the  product  of  9  xyzz  and  xz  -f  4  y*z  —  5  xyz2  —  2x*. 

Simplify 

4.  (3m+5w  +  4o-5i))(-2n3o2). 

5.  -7(5-2a+2&-3c  +  4e). 

6.  (-l)(-2m  +  5w2  +  6o-5p)(-2). 


MULTIPLICATION   AND  DIVISION  43 

7.  (a-&*-2c»-3d8)(-4a6cd)(-l). 

8.  (-2  ax*)  -(9b  +  3x-$d  +  uf-g). 
9-    (-3)(+2)(-2a  +  36  +  4c-5d). 

10.  (_5)(-l)(-2aj  + gy-32-0). 

11.  12(1+2-3  +  4-5  +  6-7  +  8). 

12.  What  are  the  two  ways  of  working  the  last  exercise  ? 
Which  of  those  two  ways  is  the  shorter  ?     Why  ? 

MULTIPLICATION  OF  POLYNOMIALS 

60.  It  is  readily  seen  that  (3  +  4)(2  +  6)  =  56.  It  might  be 
worked  also  in  this  manner: 

(3  +  4)(2  +  6)=3(2  +  6)  +  4(2+6)  =  3- 2+3- 6 +  4- 2+4-6=56. 

This  second  mode  of  procedure  is  longer  in  this  particular 
example.  It  is  given  here  because  it  illustrates  the  process 
which  must  be  followed  when  the  terms  inside  the  parentheses 
are  not  like  terms.  For  instance,  it  can  be  used  in  a  multiplica- 
tion like  the  following  : 

(x -\-2y){3x-±y)=  x(3 x -  ±y)  +2 y{3 x - ±y) 

=  x-3x  —  x-4y+2y-3x—2y-4:y 

=  3x2-4-xy  +  6xy-8y2  =  3x2+2xy-8yi. 

We  see  from  this  that  we  can  find  the  product  of  two  poly- 
nomials by  multiplying  one  polynomial  by  each  term  of  the  other 
and  then  adding  the  partial  products. 

In  written  exercises  the  work  may  be  arranged  as  follows  : 
3a;  -4y 
x  +2y 
3  x2  -  4  xy 

6xy  -  8  y2 
3x*  +  2xy-8y* 

Frequently  a  polynomial  has  some  letter  raised  to  different 
powers  in  its  terms.  Thus,  in  the  polynomial  x*  +  2x3  —  xi 
+  4  x  —  5,   all  the    terms   contain  x,    except   the   last   term ; 


44  ELEMENTARY   ALGEBRA 

moreover,  the  term  x4,  which  is  the  highest  power  of  x  in  the 
polynomial,  is  written  first.  Then  follow  the  terms  containing 
the  lower  powers  of  x,  arranged  in  descending  order.  This 
polynomial  is  said  to  be  arranged  according  to  the  descending 
powers  of  x. 

When  written  in  reverse  order,  thus,  —  5  +  4  x  —  x2  -{-  2  x3  +  x*, 
we  say  that  the  polynomial  is  arranged  according  to  the  ascend- 
ing powers  of  x.  When  polynomials  of  this  kind  are  used,  it 
is  convenient  to  arrange  them  first  according  to  the  descending 
or  the  ascending  powers  of  some  letter.  If  in  the  multipli- 
cation of  two  polynomials  both  are  arranged  according  to  the 
ascending  powers  of  some  letter,  or  both  according  to  the  de- 
scending powers,  the  partial  products  can  be  written  down  and 
added  with  much  greater  ease. 

Example.     Find  the  product  of  —  3  cc2  -|-  4  a;  —  5  +  2  a;8  and  2  x  +  x2  —  3. 
Arranging  both  polynomials  according  to  the  descending  powers  of  x, 
the  multiplication  is  as  follows  : 

2aj,-3a?+4aj  -5 

a?  +  2x  -3 

2a;5  —  3.r4  +  4ar!-  5x2 

+  4  a4 -6^  +   8  a:2 -10  a? 

-6a?  +    9  a;2 -12  a; +15 

2x>  +    o4-8ar5  +  12z2-22a;  +  15 

One  method  of  checking  a  multiplication  of  this  kind  is  to 
arrange  both  polynomials  according  to  the  ascending  powers 
of  the  letters  and  then  to  find  their  product,  thus : 
-5  +  4z-3z2  +  2arJ 

_3  +  2a;+    a;2 

+  15-12x+    Stf-Sx* 

-10*+-    8x2-6a?  +  ±xi 

-    5a?2  +  4a?-3a4+2a;5 

+  15-22z+12a;2-8z3  +    xi  +  2x5 

The  two  products  are  the  same,  except  that  one  is  arranged 
according  to  the  ascending  powers  of  x,  and  the  other  according 


MULTIPLICATION  AND  DIVISION  45 

to  the  descending  powers.  The  order  in  which  the  terms  of  a 
polynomial  are  written  does  not  affect  its  value,  hence  we  say 
that  the  two  products  are  the  same.  The  second  multiplication 
indicates  that  the  first  multiplication  was  correct. 

A  second  method  of  checking  multiplication  is  by  assigning 
some  particular  small  number  as  the  value  of  x.     Let  x  =  2, 

then  2  a:3 -3  a,-2 +  4  a; -5  =7 

x2  +  2  x  -  3 =   5 

2x5-3xi  +  4:X3—  5x* 

+  4a4-6z3  +    8^-lOa; 

-6a*>  +    9a:2 -12a; +  15 
2ar>  +    a,-4 -8  a,-3 +  12  a2 -22  a; +  15         35 

Let  x  =  2,  64  +  16  -  64  +  48  -  44  + 15  =  35. 
Hence  we  are  reasonably  certain  that  the  multiplication  is 
correct. 

EXERCISES 

61.  Multiply  together  the  following  polynomials,  and  check: 

1.  x  +  1  and  a; +  2.  9.  y  —  5  and  y3  —  2y2+  3y  —  1. 

2.  2  x  +  1  and  x  +  3.  10.  2  z  +  6  and  -  z3+  2  z2  - 10. 

3.  2  a;  —  1  and  x  —  3.  11.  3  ic  +  ?/  and  2  a4  —  3  a?  +  5. 

4.  5 x*+  x+2  and  a;  +  4.  12.  x2  +  x  and  5 a;2  +  4  x  +  2. 

5.  6cc2-2a;-2  and  —a; +  3.  13.  z2-  7  and  6 z3 -  1. 

6.  y3  —  2 y2  +  4  and  y  —  5.  \±.  x2  —  x+\  and  2 a;  +  4. 

7.  y4  —  1  and  y2—  y  —  1.  15.  a;2  —  a;  +  3  and  a;  —  $. 

8.  z2-2z+l  andz+1.  16.  z4  +  z2 +1  and  z2  — 1. 

Arrange  the  polynomials  according  to  the  descending  powers 
of  the  letter  and  multiply.  Verify  by  arranging  the  polyno- 
mials according  to  the  ascending  powers,  and  multiply  : 

17.  x  —  1  and  3^  —  4  +  23;. 

18.  6  +  z2  -  5  z  and  2  -  z2  +  3  z. 

19.  —  1  +  3  y3  +  2  y  and  6  +  5  y  —  y\ 


46  ELEMENTARY   ALGEBRA 

20.   o^-fx3-2and-5  +  2ar!-3a;  +  8x3. 
.     21.    x3  +  2  x*  -  3  x%  -  4  and  -  3  +  2  3  +  a2. 
Perform  the  indicated  multiplications : 

22.  (3x  +  y  +  z)(5x-2y-z  +  5). 

23.  [a  +  6-c][2a-2&-l-3c]. 

24.  ^  +  2^  +  0^  fa2  —  xy  +  x  +  y\. 

25.  (a2  +  y2  +  z*)(x  +  y  +  z). 

26.  (a-e  +  3/-20)2. 

27.  (a  +  b  -  c  +  d  +  e)2. 

28.  (a  +  &)(a  -  6) (a2  +  2  a&  -  62). 

29.  (x-2y)3. 

30.  (a—  y  +  z)3.  ■  31.    (a  —  b  —  c)z. 

HARDER  EQUATIONS 

62.    l.   Solve  {x  +  2)(x  -  7)  -  (as  +  5)2  =  0. 
Solution.  (x2  -  5  x  -  14)-  (x2  +  10  x  +  25)  =  0. 

Remove  parentheses,       x2  —  5  x  —  14  —  x2  —  10  *  —  25  =  0. 
Transpose,  x2  -  5  x  -  x2  -  10  x  =  24  +  14. 

Combine,  —  15  x  =  39. 

Divide  both  sides  by  -  15,  x  =  -  ||  =  -  2f. 

Cftecifc:  (-^  +  2)(- V-7)-(-¥  +  5)2  =  0. 

(-f)(-*5a)-(V)2=0- 

0  =  0. 

Solve : 

2.  (z  +  l)(z-l)-(z  +  2)(z-3)=0. 

3.  z2-(x  +  4)(a  +  l)  =  0. 

4.  (y  +  2)(y-3)  +  (y-5)(y-l)-2y*  =  0. 

5.  (22/+l)2-4(2/  +  l)2  =  0. 

6.  [>  +  l]|>-2]-0-3]0+4]  =  0. 

7.  x2  -  2  a;  +  5  -  (x  +  2)(z  -  3)  =  0. 

8-   (2/  +  3)(y-5)  +  (y  +  2)(y  +  l)-(2/  +  2)(2/-3) 

-(y-3)(^  +  4)  =  o. 


MULTIPLICATION  AND  DIVISION 


47 


9.  (x  +  l)(x  -  1)  -  (x  +  2)(x  -  2)  +  x  +  10  =  0. 

10.  (3  -  x)(2  -  x)  -  (5  +  »)(1  -  x)  -  2  x2  =  6'. 

11.  0  =  (1  -  z){z  -  2)  +  (* •+  4)(a  -  2)  +  8. 

12.  3  =  5  +  x2-  |6+aj}{aj+l|  +  5x. 

13.  (x  -  5)J  -  (*  +  2)2  =  5(*  +  3). 

14.  6  (x2  +  1)  -  (5  x  +  6) (a?  +  1)  =  (x  +  2)(x  -  2). 

15.  (x2  +  x  +  l)(x  -  5)  -  (x2  -  x  -  7)(x  -  3)  =  5(x  +  7). 

16.  x  +  10  =  —  (x  +  5)(x  -  5)  —  (4  +  x)(5  —  x). 

17.  (i/  +  3)(6^  +  7)  =  (22/-3)(3y-4)4-6. 

18.  2z-5  =  (z-  5)(z  +  4)  -  (2  z  +  l)(z  -  1)  +  z2. 

19.  5  +  (x  +  l)(x  -  l^x2  +  1)  =  (x2  +  lXx2  -  1)  +  x. 

20.  (x2  +  x  4-  2)(x  -  1)  =  (x2  -  x  -  l)(x  4- 1). 

63.  Figure  7  shows  that  a  rectangle  4  in.  long  and  3  in. 
wide  has  an  area  of  12  sq.  in.  Similar  relations  hold  when 
the  foot,  yard,  or  some  other  unit  of  length 
is  taken.  These  considerations  suggest 
the  following  definitions  : 


in.  x  m.  =  sq.  in. 
ft.  x  ft.  =  sq.  ft. 
yd.  x  yd.  x  yd.  =  cu.  yd.,  etc. 


Fig.  7 


ORAL   EXERCISES 

64.  Find  the  areas  of  rectangles  whose  dimensions  are  : 

1.  7  ft.  and  13  ft.  6.    ab  yd.  and  c  yd. 

2.  6  in.  and  9^  in.  7.   3  x  mi.  and  4  y  mi. 

3.  3|  ft.  and  4|  ft.  8.   2  ab  in.  and  7  abc  in. 

4.  5  yd.  and  x  yd.  9.   x2  yd.  and  xz  yd. 

5.  6  ft.  and  9  ft. 


48 


ELEMENTARY   ALGEBRA 


Find  the  volumes  of  rectangular  solids  whose  dimensions 


are 


10.  3  in.,  4  in.,  5  in. 

11.  5  ft.,  6  ft.,  n  ft. 

12.  2  yd.,  b  yd.,  c  yd. 

13.  I  in.,  iv  in.,  h  in. 


14.  ab  yd.,  a&c  yd.,  and  a2b  yd. 

15.  4  x  in.,  5  y  in.,  and  2  z  in. 

16.  5  y  ft.,  2  */2  ft.,  and  2  y«  ft. 

17.  x  rd.,  y  rd.,  and  xyz  rd. 


Fig.  8 


Find  the  area  of  a  square  whose  edges  are  : 

18.  6  s  ft.  20.    4  a?  in.  22.    (a  +  b)  yd. 

19.  4.5  a  in.  21.    ran2  ft.  23.    (a  +  2  ?/)  in. 

Find  the  volume  of  a  cube  whose  edges  are  : 

24.   4  a.  25.   2  62.  26.    (c  +  d)  ft. 

27.  If  the  box  shown  in  the  adjoining  figure  has  I  =  4  in., 
w  =  3  in.,  h  =  2  in., 

(a)  What  is  the  area  of  the  bottom  ? 
(&)  What  is  the  area  of  the  bottom  and  top 
taken  together  ? 

(c)  What  is  the  area  of  one  end  of  the  box  ? 
Of  both  ends  ? 

(d*)  What  is  the  total  surface  of  the  box  ? 
(e)  What  is  the  volume  of  the  box  ? 

28.  Answer  the  questions  in  Ex.  27  when  I  —  1,  w  =  2,  and 
h  is  any  number. 

29.  Answer  the  questions  in  Ex.  27  when  I  =  10  and  w  and 
h  are  any  numbers. 

30.  Answer  the  questions  in  Ex.  27  when  I,  w,  h  are  any 
numbers. 

Find  the  area  of  a  rectangle  whose  dimensions  are  as  fol- 
lows : 

31.  9,  (a  +  b).  34.    (3  -  x),  (4  +  x). 

32.  (  x  -  6),  5.  35.    (a  +  2  b),  (a  -  2  6). 

33.  (a  +  b),  (a  -  b).  36.    [5  a  -  2  6],  [2  a  +  3  6]. 


MULTIPLICATION   AND   DIVISION  49 

37.  Find  the  numerical  value  of  the  area  in  Ex.  36  when 
a  =  10,  b  =  5. 

DIVISION  BY  MONOMIALS 

65.    Just  as  in  arithmetic,  3  -*-  4  may  be  written  f ; 

so  in  algebra,  n  -~  d  may  be  written  -  • 

In  algebra,  as  in  arithmetic,  f  may  be  interpreted  in  two 
ways : 

(1)  As  an  indicated  division. 

(2)  As  a  regular  fraction,  showing  that  3  of  the  4  equal 
parts  of  a  unit  are  taken. 

The  same  is  true  of  - .     It  means  "  n  divided  by  d,"  or  "  a 

unit  is  divided  into  d  equal  parts  and  n  such  parts  are  taken." 
We  can  adopt  either  interpretation. 

We  have  learned  in  arithmetic  that  the  value  of  an  expres- 
sion (quotient  or  fraction)  is  not  changed,  when  the  dividend 
and  divisor  in  a  division  (or  the  numerator  and  denominator 
of  a  fraction)  are  both  multiplied  or  both  divided  by  the  same 
number. 

That  is,  |  =  |  =  f 

This  principle  is  used  in  simplifying  the  dividend  and 
divisor,  or  in  reducing  a  fraction  to  its  lowest  terms. 

Thus,  dividing  by  5  a   both  terms  of  the  fraction  —  — — — ,  we  obtain 

..  ,        Abe 

its  equal,  —  — — -  • 
a 

The  rule  of  signs  must  be  followed  carefully.  In  division,  as  in  multi- 
plication, "  like  signs  give  plus,  unlike  signs  give  minus." 

We  know  (§  13)  that  25  =  2  •  2  . 2 . 2  •  2,  and  23=  2  •  2  . 2. 

25      2  •  2  •  2  •  2  •  2 
Hence  |  =        2>2>2      =  V  or  2-. 

_,..,.  a5     a-  a-  a  •  a  •  a        ,        u 

Similarly,        —  = =  a2  or  a6-3- 

■"        a3  a-  a  ■  a 


50  ELEMENTARY   ALGEBRA 

This  gives  the  following  very  important  rule : 
The  exponent  of  any  letter  in  the  quotient  is  equal  to  its  ex- 
ponent in  the  dividend  minus  its  exponent  in  the  divisor. 

It  is  thus  seen  that  in  division  we  subtract  the  exponents  of  any  letter, 
while  in  multiplication  (§  56)  we  add  its  exponents. 

In  general,  in  multiplication,  am  •  an  =  am+n ; 

.  .        am 

in  division,  —        =  am~n- 

'  an 

ORAL  EXERCISES 

66.  Perform  the  indicated  divisions : 

x     12  abc     24afy     36  abtf     Scdx2 
Aab        6xy       12  aby'    12  dx 
-a2     —  c3      +  z*      -3m3      +5a^ 
—  a  '       c  z2    '    +  3m  '      —  x3 

abb      —ab*c      —  4  arty4      +  9  m2n5 
ab '    —  2  ab3 '    +  4  x*y2'    —  3  m2n2 

67.  EXERCISES 

1.  -  124  aWc5  -=-  12  aW. 

2.  -  25  ofy7*9  h-  15  x2y7z\ 

3.  236  m6n2o5  ^- 24  m%6o8. 

4.  +  50  btfcPe6  +  ( -  60  fcWe3) . 

Reduce  the  following  fractions  to  the  lowest  terms  : 

-  28  pYr9  +  72  aWc12  —  64  Pm10tt10 

+  56pVrV'  —  44aWd7'      '    -196Pm6n10o5' 

Let  >4  be  the  area  of  a  rectangle,  L  its  length,  W  its  width. 
Find  the  second  dimension  in  the  following  rectangles  : 

8.  A  =  56  a2  sq.  ft.,  L  =  8  a  ft. 

9.  A  =  136  afy2  sq.  in.,        L  =  12xy  in. 

10.  .4  =  365mVo6,  TT=  25  m3n5o2. 

11.  ^4  =  256&c8d3,  W  =72  We'd2. 

12.  .4  =  1.25 x4y8,  Z  =  .05xy\ 


MULTIPLICATION  AND   DIVISION  51 

MULTIPLICATION  OF  FRACTIONS 

68.  Figure  9  shows  that  £  of  f  of  an  inch  is  f  of  an  inch ; 
this  recalls  to  mind  the  arithmetical  rule  for  the  multiplication 
of  fractions :  Multiply  the  numerators  to- 
gether/or a  new  numerator,  and  the  denomi- 
nators together  for  a  new  denominator. 

The  process  is  the  same  in  algebra. 

It  is  easiest,  as  a  rule,  at  first  merely  to 
indicate  the  multiplication,  then  to  divide  both  numerator  and 
denominator  by  every  factor  common  to  them. 
My  a262 

Thus,        -«2*S times  ***  J&*£l*f#  =  8aW . 

C-  1)  c      3 

EXERCISES 

69.  Eind  the  product  of  the  following  fractions : 
3ab     12  ed  -  2  a5     -  3  g«y 

16zm;2     85  aHb  g  ±7_£  x  _8jL  . 

15  a*  '    64zw>  '  .  -24       -14* 

1          64  a^z3  6  -  24  a2bc*      -  2  m3n2 

8ayz'        a&c    '  "  -36mV       14  aW  * 

Since  in  division  "unlike  signs  give  minus,"  it  is  evident  that  the 
following  equalities  hold : 

~b~~~^b~~b' 

That  is,  the  —  sign  may  be  in  front  of  the  fraction,  or  before  a  factor 
of  the  numerator,  or  before  a  factor  of  the  denominator,  and  there  is  no 
change  in  the  value  of  the  fraction. 

?     -  25  ah*      +  2  ^2e3  -  afyV      -  atfc? 

8aV  5a4c2   '  *    —  a3b*cb      -x*yz° 

27  mbm        3  rW  ,_        ifWM  5thy*w 


l. 


2. 


8.     .„  I .  x     .....  ..  io 


12  s¥        —  9fc8Z4m2  —  l5Pyw2      +4/2^2/<4* 


52 


ELEMENTARY   ALGEBRA 


If  a  fraction  -  is  to  be  multiplied  by  an  integer  c,  it  is  easiest  to  write 
b 

for  c  its  equal  -,  and  then  to  multiply   -  by  -,  by  the  rule  for  the 
1  6  1 

multiplication  of  fractions. 

4a62cd3 


11.    24a2c3e3  times 


6e2fg 


13.   32  63cxd6  times 


2ert5 
6W" 


12. 


30  7iH% 


times  3xW.  14.    £ — times  —5  hmtn3. 

21  afy2z5  -3mVoJ 


DIVISION  OF  FRACTIONS 

70.    In  algebra,  as  in  arithmetic,  a  fraction  is  divided  by 
another  "  by  inverting  the  divisor  and  multiplying." 

This  rule  may  be  proved  as  follows :  Let  the  quotient  of  -  ■+■  -  be  x. 

b     a 

The  dividend  is  -  ;  the  divisor  is  - . 
b  d 


Since 


divisor  x  quotient  =  dividend,  we  have, 

c  a 

-  x  x  =  ~,  or 
d  b 


equation  by  bd  gives 

the  lowest  terms, 
coefficient  of  x,  we  get 


—  =  -.     Multiplying  both  sides  of  the 
d      b 

bcdx  =  qbd^  or^  reducing  the  fractions  to 
d  b 

bcx  =  ad.     Dividing  both    sides    by    the 

„     ad 

x  =  — 
be 


This  value  of  x  is  the  required  quotient.     One  sees  at  once  that  a  quick 
way  of  getting  this  answer  is  to  invert  the  divisor  and  multiply. 


ORAL  EXERCISES 
71.    Perform  orally  the  following  divisions  : 


1        3_s_l 

1.  ¥— T 


4.   14  + 24-. 


6.    b  +  «. 
c     e 


mn  .  pq 

pq      m3 

4a2  .  Sab 
b     '     c 


MULTIPLICATION  AND   DIVISION  53 


In  divisions  like  a  -=-  -  or  -  -4-  a,  it  is  easiest  to  write  -  in  place  of  a  ; 
c        c  I       * 


then  proceed  as  in  the  previous  exercises. 


9. 

a 

°  *  b' 

11.    5- 

m                            6c  .  3 

10. 

X 

y 

-  \ 

.a                             2a  .  2 
'6*                       "36*3' 

72. 

WRITTEN 

EXERCISES 

l. 

14  as2d?  . 
25  x2?/ 

28  a2s2d 
125  xy2 ' 

3      72  mW  .  288  mnWp 
-91H10  "       98  fc5^6 

1 

256  e«r¥ 

128  eWy 

-  135  <f/y  .  155  d/3a/i4 

36  a&V 

'  144  a362c2d" 

84  xyz            91  xyz2 

ORAL  PROBLEMS 
73.    1.    How  many  25^  pieces  make  $4? 

2.  How  many  25  ^  pieces  make  $  d  ? 

3.  A  man  "steps  off"  the  length  of  a  playground  and  finds  it 
to  be  125  paces.     How  many  feet  is  this,  if  his  pace  is  3  ft.  ? 

4.  If  a  man's  pace  is  3  ft.,  what  distance  do  p  such  paces 
measure  ? 

5.  A  man  finds  the  width  of  a  lot  to  be  a  lengths  of  a  cer- 
tain rod  which  is  b  ft.  long.     How  wide  is  the  lot  ? 

6.  The  water  in  a  canal  flows  at  the  rate  of  -  mi.  an  hour. 

h 
In  what  time  will  it  flow  -  mi  ? 

k 

7.  An  aviator  flies  at  the  rate  of  95  mi.  an  hour.    How  long 

will  it  take  him  to  go  -  of  a  mile  ? 
s 

8.  A  girl  cuts  off  two-thirds  of  a  ribbon  2\  yd.  long.     How 
many  feet  of  ribbon  does  she  cut  off  ? 

9.  A  woman  gets  the  -  part  of  a  supply  of  coffee  weighing  - 

a  c 

pounds.     How  many  pounds  does  she  receive  ? 


54  ELEMENTARY   ALGEBRA 

DIVISION  OF  POLYNOMIALS  BY  MONOMIALS 

74.  This  is  easily  done  by  dividing  each  term  of  the  poly- 
nomial by  the  monomial  and  connecting  these  partial  results  by  the 
proper  signs. 

Thus,  (15  x2y2  -6xy3  +  12xy)^-(3  xy) 

=  — — - —  H -  =  5  xy  —  2  y2  +  4,  the  quotient. 

Sxy        oxy       oxy 

Again,  (a2b3  -  3  a26*  -  5  a363)-H  5  a262 

a263      3a264     5a363     b     3  62       ,    ., 

= = ao,  the  answer. 

oa2b2     5a2b2     5a2b2     5       5 

The  correctness  of  the  answer  can  be  tested  by  multiplying  the  quotient 
by  the  divisor.    The  resulting  product  should  be  equal  to  the  dividend. 

EXERCISES 

75.  Perform  the  indicated  divisions  and  check  your  results : 


1 

8  x2  —  6  xy 

6  x>y2  —  4  afy5  +  8  a%2 

2x 

4  arty2 

2 

12  a3b2  + 6  a2b2 

36r4.s3+48r3s3-24r2s3 

3a2b2 

12^ 

3 

9  mk*  -  12  m2k* 

8  a2z2  -  10  a*/z2  -  8  afyV 

Smk* 

6  a;2?/2^2 

4. 

15  b3c2  -  20  Wed 
15  b2c 

3  miV  +  5  w2z7  +  7  w^z3 
4iuV 

5. 

7/^  +  14  h2l* 
7hP 

—  5  abc  —  6  c^&c2  +  8  a2bc 
—  a&c 

6. 

24  - 12  as3 
12 

,  0     c2d?e  —  4  cd?e2  +  &dz& 

—  cd?e 

13     20a^" 

■24afy2 +  12ay 
-\2x*y 

..     18  ab2<? 
14.     

-  12  aW  +  24  aW 

12  a&V 


MULTIPLICATION  AND  DIVISION  55 

DIVISION  OF  POLYNOMIALS  BY  BINOMIALS 
76.    An  inspection  of  the  following  multiplication  will  sug- 
gest some  of  the  steps  to  be  taken  in  the  inverse  operation  — 
Division.  4x2  +  3x.+  5 

x-2 

Ax3  +  '3x2  +  5x 

-8a-2-6a;-10 
4  x3  —  5  x2  —    x  —  10 

If  we  are  required  to  divide  4  x3  —  5  x"-  —  x  —  10  by  x  —  2, 
then  the  quotient  which  must  be  the  other  factor,  4  x2  -f  3  a;  +  5, 
can  be  found  as  follows: 

Dividend  Ax3  —  5x2  —  x—10  x  —  2     Divisor 

1st  partial  product  4  x3  —  8  x2  4#2.+  3x  -f  5  Quotient 

1st  remainder  +  3  x2  —  x  —  10 

2d  partial  product         +  3  x1  —  6  x 

2d  remainder  +  5  x  —  10 

3d  partial  product  +  5  x  —  10 

Last  remainder  0 

Notice  that  both  the  given  dividend  and  divisor  are  arranged  accord- 
ing to  the  descending  powers  of  the  leading  letter  x.  The  ascending 
powers  would  have  answered  equally  well.  It  is  a  very  great  conven- 
ience to  have  the  dividend  and  divisor  arranged  either  both  according  to 
the  ascending  or  both  according  to  the  descending  powers  of  a  leading 
letter.  If  they  are  not  so  arranged,  then  they  should  be,  before  the 
division  is  begun. 

In  the  division  above,  notice  the  following : 

(1)  The  first  term  in  the  quotient  4  x2  is  obtained  by  dividing  the 
first  term  in  the  dividend  Ax3  by  the  first  term  in  the  divisor  x. 

(2)  The  first  partial  product  is  the  product  of  the  first  term  in 
the  quotient  4  x2  arid  the  divisor  x  —  2. 

(3)  The  first  remainder  3  x2  —  x  —  10  is  obtained  by  subtracting 
this  partial  product  from  the  dividend. 

The  —  10  is  not  needed  in  the  second  step  of  the  division 
and  may  therefore  be  omitted  until  the  third  step. 


56  ELEMENTARY   ALGEBRA 

The  remaining  steps  of  the  division,  should  there  be  any,  are 
repetitions  of  steps  above. 

77.    Explain : 

(1)  How  is  the  term  +  3  a  of  the  quotient  found  ? 

(2)  How  is  the  second  remainder  found  ? 

(3)  How  are  the  +  5  in  the  quotient  and  the  third  partial 
product  found  ? 

(4)  How  can  you  check  your  division  ? 

In  many  cases  the  division  does  not  come  out  exact  and  there  will  be 
a  final  remainder.  ,In  exercises  1-23  which  follow,  the  divisions  will 
come  out  exact. 

Divide  9  -  9  a  +  8  a2  -  7  a3  +  2  a4  by  3  -'2  a. 

2x 


9-9a  +  8a2- 

-  7  x3  +  2  x* 

9-6a 

-Sx  +  Sx1* 

-3a +  2  a2 

'       +  6  a2  - 

-7a3 

+  6a2- 

-4a3 

-3o3  +  2o4 

-3a3  +  2  x* 

3-a  +  2o2-o3 


EXERCISES 

78.    Divide: 

1.  o2  +  2a  +  l  by  05  +  1.  5.  o3-2o2+2a— lbya-1. 

2.  y2  -f  4  y  +  4  by  y  +  2.  6.  w3  +  1  by  w  +  1. 

3.  x2  -  a  —  6  by  x  -  3.  7.  c3  —  1  by  c  —  1. 

4.  z2  -  7  z  +  10  by  z  -  5.  8.  a4  -  16  by  a  -  2. 

9.   6^  + a3 -15  a2 +  38  a -30  by  x-1. 

10.  4  +  28 x  +  29a2  -30^  +  25a4  by  5a  +  2. 

11.  y>  +  2y*-3y3-y2-2y  +  3byy  +  3. 

12.  a2  —  62  by  a  —  &.  (  14.   a2  +  2  xy  +  y2  by  a  +  y. 

13.  a2  —  y2  by  a  +  y.  15.   a2  —  2  a?/  +  y2  by  a  —  y. 


MULTIPLICATION  AND   DIVISION  57 

16.  ra3  -f  n3  by  ra  +.n.  20.  c6  —  (f  by  c8  —  d3. 

17.  m3  —  n3  by  m  —  n.  21.  a6  —  66  by  a3  +  &3. 

18.  x6  —  y6by  x  —  y.  22.  x6  —  y6  by  x  —  y. 

19.  a6  —  66  by  a2  —  b\  23.  a5  +  yh  by  a  +  y. 

DIVISIONS  INVOLVING  REMAINDERS 

79.    11  h-  3  gives  the  quotient  3  and  the  remainder  2 ;  the  division  is 
not  exact ;  the  answer  is  3  and  f .     We  see  that 

dividend  +  divisor  =  quotient  +  remainder  . 

divisor 

Divide  x5  +  3x3-2x2  +  x-2byx8-2. 

x5  -jr  3  x3  -  2  x2  +  x  -  2     1  x8  -  2 
x5  -  2  x2  x2  +  3 


+  3  x3  +  x  -  2 

+  3  x8  -  6 


+  x  +  4,  the  remainder. 

x  +  4 

Hence  the  quotient  is         x2  +  3  + -. 

x8  —  2  , 

Check :  dividend  =  quotient  x  divisor  +  remainder. 
EXERCISES 

80.   Divide: 

1.  x3  +  2  x2  +  7  x  —  5  by  x  —  9. 

2.  y1  -  4  f  +  .5  y8  -  3  y  +  12  by  y>-  +  4. 

3.  z5  -  16  z3  -  6  z2  +  3  2  +  5  by  z2  -  4  z. 

4.  ttf8  4- 1  by  w"1  +  1. 

5.  a4  +  3  a2  +  9a  -  10  by  a3  -  7. 

6.  m5  —  5  ra4  —  8  ?/i3  +  4  ra  -f-  8  by  m3  + 1. 

7.  r16  +  1  by  r*  +  1. 

8.  s16  -  1  by  s5  -  2. 


CHAPTER   IV 

PROPORTION 

81.  A  proportion  expresses  the  equality  of  two  common 
fractions. 

Thus,  ||  and  §  are  equal  to  each  other ;  hence,  f|  =  §  is  a  proportion. 

A  common  fraction  may  be  said  to  indicate  the  division  of  the  number 
in  the  numerator  by  the  number  in  the  denominator.  When  so  con- 
sidered, the  fraction  is  frequently  called  a  ratio.  The  ratio  -  is  often 
written  a :  b. 

Since  a  proportion  is  a  special  type  of  equation,  it  may  be  treated  like 
an  equation.  If  one  of  the  terms  of  either  fraction  in  the  proportion  is 
unknown,  the  rules  for  the  solution  of  equations  enable  us  to  find  the 
value  of  the  unknown  term. 

x      3 

1.  Solve  the  proportion  —  =  -. 

1  _      o 

Multiplying  both  sides  of  the  equation  by  numbers  which  will  remove 
the  denominators,  that  is,  multiplying  both  sides  by  12  and  by  5,  we  get 
5  •  yl  •  x  _  ft  .  12  •  3 

Or  5  x  a  36. 

Divide  both  sides  by  5,  x  =  *£  =  7$,  the  answer. 

2.  Solve  the  proportion =  -. 

c  d 

Multiplying  both  sides  by  c,  and  also  by  <f, 
cd(a  —  b)  _cdx 
c  d 

Or  d(a  —  b)  =  ex. 

Change  sides,  for  convenience,      ex  a  d(a  —  b). 

Divide  both  sides  by  c,  x  =  "(q  7  ft);  the  answer. 

68 


PROPORTION  59 


EXERCISES 

82.    Solve  the  following  proportions : 


1. 

x     23 
6      3  ' 

8. 

x  _mn 
12_X' 

2. 

.T_19 

5      3  ' 

9. 

—  c_     y 
16       -32 

3. 

S_x 
9     3' 

10. 

5_10 
x     25* 

4. 

!0_  y 

7       -4 

11. 

x        3 

c  —  d     4 

5. 

7       x 
8~3s' 

12. 

a      n 

6. 

10.5 
6 

a: 
8* 

13. 

'  _  x  . 
20~m' 

7. 

a  _  z 
5~15' 

14. 

-6      s 

15.  a;:  5  =  19:  3. 

16.  -10:7  =  ?/:4. 

17.  10.5:6  =  -a;:8. 

18.  x  :  12.5  =  —  mn  :  4. 

19.  5.5:  a;  =10.7:  25.4. 

20.  —  x:  a  =  m:n. 

21.  —  r:  —6.8  =  y.s. 


PROBLEMS  IN  PROPORTION 

83.  1.  If  5  acres  of  land  are  worth  $875,  what  are  7  acres 
of  land  of  the  same  quality  worth  ? 

Here  the  number  of  acres  is  said  to  be  in  the  "same  ratio"  as  or 
"  directly  proportional "  to  the  cost.    That  is, 
5  acres  _  $875 
7  acres       <$x 

Since   5acres  =  5  and  ^^  =  — ,  the  original  proportion  can  be  written 
7  acres      7  |  z         x 

in  the  simplified  form,  5  _  875 

l~~x" 

We  get  x  =  1225.     Hence,  the  answer  is  $1225. 

2.  If  11  men  can  build  a  wall  in  80  days,  how  long  would  it 
take  14  men  to  do  the  same  work  ? 

Here  the  larger  the  number  of  men,  the  less  the  time  for  doing  a  given 
piece  of  work;  the  number  of  men  is  not  in  the  "same  ratio"  as  the 


60  ELEMENTARY   ALGEBRA 

times  needed  to  complete  the  work,  but  in  the  "  inverse  ratio."  That  is, 
the  number  of  men  is  "inversely  proportional"  to  the  time  required  to 
finish  the  work.     We  have,  therefore, 

11  men      x days  •      , 

—  = i— ,  or  more  simply, 

13  men     80  days 

11  _  x_ 

13     80* 

Solve  this  proportion. 

3.  If,  in  a  broad  jump,  a  boy  can  jump  15  ft.  when  he  takes 
a  running  start  of  20  ft.,  how  far  can  he  jump  with  a  running 
start  of  400  ft.  ? 

Is  the  distance  a  boy  can  jump  either  "directly"  or  "inversely" 
proportional  to  the  running  start  ? 

Can  this  example  be  worked  by  proportion  ? 

In  working  problems  we  must  consider  the  following  : 

(1)  Can  the  problem  be  worked  by  proportion ;  that  is,  do  the  quan- 
tities named  in  the  problem  vary  in  either  direct  or  in  inverse  ratio  ? 

(2)  If  they  do,  determine  whether  the  variation  is  in  "direct  ratio" 
or  in  "inverse  ratio." 

It  should  be  observed  that,  while  a  variation  may  not  be  strictly  in 
"direct"  or  in  "inverse  ratio,"  the  variation  may  be  so  nearly  in  that 
ratio  that  the  method  of  proportion  yields  answers  that  are  sufficiently 
accurate  for  practical  purposes. 

For  example,  it  does  not  necessarily  follow  that  4  men  can  do  a  piece 
of  work  in  half  the  time  that  2  men  can  do  the  work.  The  4  men  may 
not  be  equally  efficient  or  4  men  may  be  in  each  other's  way,  so  that  4 
men  could  not  work  as  efficiently  as  2  men  can.  But  in  many  such  cases 
the  differences  are  so  slight  as  to  be  negligible. 

4.  Find  two  numbers  in  the  ratio  of  4  to  5  whose  sum  is 
207. 

Let  the  two  numbers  be  4  a;  and  5  x. 

5.  Find  two  numbers  in  the  ratio  of  3  to  8  whose  sum  is 
-275. 

6.  Find  two  numbers  in  the  ratio  2  to  —  9  whose  sum  is 
-217. 

7.  Find  two  numbers  in  the  ratio  of  —2:5,  such  that  the 
first  number  minus  the  second  yields  —  133. 


PROPORTION 


61 


8.  Two  sums  of  money  bear  interests  which  are  in  the 
ratio  of  6  to  7.  Find  each  interest,  if  the  two  interest  sums 
together  amount  to  650  dollars. 

9.  Two  sums  of  money,  at  the  same  time  and  rate,  bear 
interests  which  are  in  the  ratio  of  3  to  4.  What  is  the  larger 
sum,  if  the  smaller  is  $  3360  ? 

10.  Two  equal  sums  of  money,  both  at  the  same  rate,  bear 
interests  which  are  to  each  other  as  10  is  to  13.  If  the  time  for 
the  lesser  interest  is  5  years  and  5  months,  what  is  the  time  for 
the  other  ? 

11.  What  is  the  height  of  a  flagstaff  which  casts  upon  level 
ground  a  shadow  of  84  ft.,  if  at  the  same  moment  a  vertical  rod, 
10  ft.  long,  casts  a  shadow 
7  ft.  long  ? 

As  shown  by  Fig.  10,  two 
triangles  are  formed  which 
have  their  sides  parallel,  re- 
spectively. Such  triangles 
have  the  same  shape,  but  not 
the  same  size.  It  is  shown 
in  geometry  that  such  tri- 
angles have  their  correspond- 
ing sides  proportional.  In 
this  case  the  horizontal  sides 
are  to  each  other  as  the  verti- 
cal sides  and  as  the  slanting 
sides.  In  our  problem  the  slanting  sides  are  not  used  in  forming  the 
proportion.     We  form  the  proportion  thus: 

Horiz.  side  of  small  tri.  _  Vert,  side  of  small  tri. 
Horiz.  side  of  large  tri.  —  Vert,  side  of  large  tri. 

84  _  x  - 
x  being  the  required  height  of  the  flagstaff. 

12.  A  pole  for  wireless  telegraphy  casts  a  shadow  of  105  ft. 
upon  level  ground  at  the  same  time  that  a  vertical  rod,  12  ft. 
long,  casts  a  shadow  7  ft.  long.     Compute  the  length  of  the  pole. 


7  ft. 


84ltT 


Fig.  10 


62  ELEMENTARY   ALGEBRA 

13.  By  the  method  just  explained  determine  the  height  of 
the  flagpole  of  your  school,  or  of  some  similar  high  object. 

14.  At  the  moment  when  the  sun  sets  behind  a  mountain 
peak,  a  vertical  rod  12  ft.  high  casts  a  shadow  of  27  ft.  on  level 
ground.  If  the  mountain  is  known  to  tower  7000  ft.  above  the 
plain,  what  is  the  distance  in  a  horizontal  line  to  the  center  of 
the  mountain  ? 

15.  Concrete  is  made  by  mixing  cement,  sand,  gravel,  and 
water.  A  1 :  2  :  4  concrete  is  made  up  of  one  volume  of  cement, 
twice  that  volume  of  sand,  and  4  times  that  volume  of  stone. 
The  volume  of  the  mixed  cement  is  less  than  the  sum  of  the 
volumes  of  the  parts,  because  of  the  open  spaces  between  the 
stones  which  must  be  filled  in  making  the  concrete.  One  cubic 
foot  of  concrete  contains : 

for  1:2:4  concrete :  .22  cu.  ft.  cement,  .44  cu.  ft.  sand,  .88 
cu.  ft.  gravel. 

for  1 :  2\  :  5  concrete  :  .19  cu.  ft.  cement,  .475  cu.  ft.  sand, 
.95  cu.  ft.  gravel. 

In  the  building  of  a  silo,  1125  cu.  ft.  of  concrete  is  needed,  f 
of  which  must  be  1 :  2  :  4  concrete,  the  rest  1 :  2^  :  5  concrete. 
How  many  cubic  feet  of  cement,  of  sand,  and  of  gravel  are 
needed  for  making  the  1:2:4  cement  ? 

Let  the  required  number  of  cubic  feet  of  cement,  sand,  and  gravel  be, 
respectively,  22  x,  44  x,  and  88  x. 

16.  In  Ex.  15,  how  many  cubic  feet  of  cement,  of  sand,  and 
*  of  gravel  are  needed  for  making  the  1 :  2\  :  5  cement  ? 

17.  A  contributes  $  2000  to  a  speculation,  B  $  3500,  C  $  1500. 
The  total  profits  are  $  725.  What  amount  of  the  gain  ought 
each  to  receive  ? 

18.  Find  a  number  such  that  its  excess  over  8  bears  the  same 
ratio  to  its  excess  over  9  as  the  number  itself  does  to  its  excess 
over  2. 


PROPORTION  63 

19.  Two  pounds  of  tea  cost  as  much  as  3  lb.  of  coffee,  and  as 
much  as  5  lb.  of  butter.  The  cost  of  1  lb.  of  tea,  1  lb.  of  coffee, 
and  |  lb.  of  butter  is  $  1.40.  Find  the  cost  per  pound  of  each 
article. 

GRAPHIC  REPRESENTATION  OF  TEMPERATURE 

84.  If  we  desire  information  on  the  changes  of  temperature 
for  any  period  of  time,  we  may  take  readings  of  a  thermometer 
at  certain  intervals  of  time.  We  may  take  hourly  readings  of 
the  outdoor  temperature  during  the  afternoon  and  evening  of 
a  winter  day  and  obtain  data,  in  degrees  Fahrenheit,  perhaps 
as  follows : 

Noon,  20°.  7  p.m.,  15°. 

1  p.m.,  21°.  8  p.m.,  9°. 

2  p.m.,  22°.  9  p.m.,  5°. 

3  p.m.,  21.5°.  10  p.m.,  0°. 

4  p.m.,  20°.  11  p.m.,  -  2.° 

5  p.m.,  17°.  12  p.m.,  -  3°. 

6  p.m.,  16°. 

These  readings  convey  an  idea  of  the  changes  of  tempera- 
ture during  the  12  hours  following  midday.  We  see  that  the 
maximum  temperature  was  at  2  p.m.,  the  minimum  at  mid- 
night.    The  temperature  dropped  most  rapidly  between  7  and 

8  P.M. 

These  relations  can  be  seen  much  more  easily,  if  the  changes 
of  temperature  are  shown  graphically.  Draw  a  horizontal  line, 
as  in  Fig.  11,  and  let  successive  equal  intervals  from  o  along 
the  line  ox  indicate  the  successive  hours  after  midday.  We 
mark  in  this  way  the  hours  1,  2,  3,  etc. 

At  each  of  these  points  we  draw  a  line  perpendicular  to  ox 
to  represent  the  temperature  at  that  particular  time.  If  the 
temperature  is  +,  we  measure  the  distance  from  ox  upward, 
but  if  the  temperature  is  — ,  we  measure  the  distance  from  ox 
downward.     If  1°  is  represented  by  the  length  of  one  of  the 


64 


ELEMENTARY   ALGEBRA 


divisions  up  or  down,  then  the  midday  temperature  is  repre- 
sented by  20  spaces  from  ox  up,  while  —  3°  at  midnight  is 
measured  from  ox  by  3  spaces  down.     If  we  connect  the  ends 


v 

20" 

00 

4>    in 
U    10 

bo 

Q 

o 

X 

u 

M 

> 

} 

i 

1 

i 

* 

- 

1 

\ 

<) 

l\ 

1 

1 

1ST 

V 

H 

"b 

Fig.  11 


of  the  lines  thus  drawn,  we  obtain  a  broken  line  which  shows 
to  the  eye  the  changes  in  temperature.  This  broken  line  is 
the  graphic  representation  of  temperature.  It  is  called  a  graph 
or  a  diagram.  Drawing  such  a  graph  or  diagram  is  called 
plotting  it ;  locating  a  point  is  called  plotting  the  point. 

How  can  you  tell  from  the  graph  the  time  when  the  tempera- 
ture was  highest  ?  The  time  when  it  was  lowest  ?  The  time 
when  the  temperature  changed  least  ?  The  time  when  the 
temperature  changed  most  rapidly  ?  What  was  the  drop  in 
temperature  between  2  p.m.  and  9  p.m.  ?  Between  7  p.m.  and 
10  p.m.  ?  Between  8  p.m.  and  12  p.m.  ?  Between  1  p.m.  and 
11p.m.? 


PROPORTION  65 

Care  should  be  taken,  in  plotting,  to  use  a  scale  not  too 
small  and  yet  small  enough  so  that  all  the  statistics  can  be 
used. 

EXERCISES 

85.  1.  Plot  the  curve  of  temperatures  for  the  next  12  hours 
the  observed  temperatures  being  as  follows : 

12  p.m.,  -  3°.  7  a.m.,  -  4°. 

1  A.M.,    -  4°.  8  A.M.,  0°. 

2  A.M.,   -  4°.  9  A.M.,   +  3°. 

3  A.M.,    -  5°.  10  A.M.    +  7°. 

4  A.M.,   -  5°.  11  A.M.    4- 10°. 

5  a.m.,  -  6°.  12  Noon,  +  15°. 

6  A.M.,    -  6°. 

2.  Find  the  rise  in  temperature  between  5  a.m.  and  11  a'm. 

3.  At  what  time  was  the  rise  in  temperature  most  rapid? 

4.  What  was  the  difference  in  temperature  between  12  p.m. 
and  12  noon  ? 

5.  Plot  the  curve  of  mean  temperatures  for  1913,  Denver, 

Col. 

Jan.  30°.  July  72°. 

Feb.  22°.  Aug.  73°. 

Mch.  38°.  Sept.  59°. 

Apr.  49°.  Oct.  46°. 

May  58°.  Nov.  44°. 

June  67°  Dec.  23°. 

6.  Plot  the  rise  and  fall  of  the  tide  on  the  coast  of  Southern 
California,  Aug.  2,  1913. 

At  3  :  20  a.m.,  -  1.2  ft. 
9 :  50  a.m.,  -  4.2  ft. 
2:50  p.m.,  1.7  ft. 
9:00  p.m.,  7  ft. 

r 


66 


ELEMENTARY   ALGEBRA 


FLUCTUATIONS   IN  THE  PRICE  OF  COAL 
ORAL   EXERCISES 

86.   Figure  12  shows  the  changes  in  the  price  of  anthracite 
coal  in  the  United  States  from  1860  to  1914. 

The  years  are  shown  on  a  horizontal  line,  the  cost  on  a  ver- 


rV-V 


^ 


Year 
Fig.  12 


PROPORTION  67 

tical  line.     For  instance,  in  1890  the  price  per  ton  is  seen  to 
be  $3.90;  in  1910,  $4.80. 

1.  What  effect  had  the  Civil  War  upon  the  price  of  coal  ? 

2.  In  what  year  was  the  price  $4.50? 

3.  Xame  the  highest  price  shown. 

4.  Name  the  lowest  price  shown. 

5.  How  many  tons  of  coal  could  be  bought  in  1893  for 
$57.20? 

6.  What  was  the  difference  in  the  price  per  ton  in  1874  and 
1861? 

7.  What  was  the  difference  in  the  price  of  100  tons  in  1866 
and  1899  ? 

DRAWING   EXERCISE 

87.  1.  Draw  a  diagram  showing  the  fluctuations  in  the  price 
of  iron  per  ton  during  the  years  1870  to  1910,  the  data  being 
as  follows : 

1885,  $18.  1900,  $20. 

1887,  $21.  1902,  $16. 

1889,  $18.  1903,  $21. 

1890,  $18.50  1904,  $19. 

1894,  $13.  1905,  $13. 

1895,  $13.  1906,  $20. 

1898,  $12.  1907,  $23. 

1899,  $19.  1910,  $17. 

2.  Make  a  table  of  varying  quantities  in  your  own  experience 
and  draw  the  graph.  Select  the  number  of  pupils  in  your 
school,  the  amount  of  money  you  have  spent  in  successive 
weeks  or  months,  the  amount  of  rainfall  in  successive  months, 
the  number  of  immigrants  coming  to  America  in  different 
years,  or  some  similar  data  which  you  are  able  to  obtain. 

88.  From  these  graphs  we  can  see,  at  a  glance,  the  variation 
in  one  thing  as  another  changes.  For  instance  in  Fig.  12  there 
is  a  change  in  the  price  of  coal  per  ton  as  the  time  increases. 


1870, 

$33. 

1872, 

$44. 

1874, 

$30. 

1877; 

$19. 

1878, 

$18. 

1880, 

$28. 

1881, 

$25. 

1882, 

$26. 

68 


ELEMENTARY   ALGEBRA 


These   quantities   are  called   variables  ;   the  value   of   one 
variable  depends  upon  the  value  of  the  other. 


89.   This  idea  can  be  carried  over  to  the  relation  between 
two  unknown  quantities  expressed  by  an  equation. 
For  instance,  in  x  +  y  =  7 : 


if  x  =  3, 

then 

3  +  2/  =  7, 

hence   y  =  4 

if  x  =  2, 

then 

2  +  y  =  7, 

hence  y  =  5 

if  x  =  1, 

then 

1-Hf-T, 

hence   y  =  6 

if  x  «  0, 

then 

0  +  y-7, 

hence   y  =  1 

if  x  =  — 

1, 

then 

-l+y-T, 

hence   y  =  8 

if  x  =  — 

4, 

then 

-  4  +  y  =  7, 

hence   y  =  9. 

As  a;  increases,  ?/  decreases ;  and  as  a;  decreases,  y  increases. 

Thus  x  and  y  are  two  variables,  the  value  of  y  depending 
upon  what  the  value  of  x  may  be,  and  the  value  of  x  depending 
upon  the  value  of  y;  thus  x  is  a  function  of  y  and  y  is  a  function 
of  a;. 


90.  Let  us  make  a  table,  assuming  ten  values  of  x  and  com- 
puting the  corresponding  values  for  y  in  the  equation  x  —  y  =  5. 
The  result  is  the  following  table : 

We  count  off  values  of  x  on  the  horizontal  line 
OX  (called  the  x-axis)  and  the  corresponding 


X 

-y  = 

5 

X 

y 

Point 

6 

i 

^ 

5 

0 

5 

4 

-  i 

c. 

3 

-2 

D 

2 

-3 

E 

1* 

-H 

F 

1 

-4 

0 

0 

-5 

H 

-  1 

-6 

I 

-2 

-7 

J 

v 

Ja 

X 

O 

R 

•Vc 

D 

gfrii 

<^J 

! 

ri 

'I 

/ 

J 

Fig.  13 


PROPORTION  69 

values  of  y  on  the  vertical  line  OY  (called  the  y-axis),  counting  to  the 
right  of  the  origin  0  for  +  values  of  x  and  to  the  left  for  —  values  of  x  ; 
up  for  +  values  of  y,  and  down  for  —  values  of  y. 

This  becomes  plainer,  if  we  plot  each  of  the  points  in  Fig.  13.  The 
first  point,  x  =  6  and  y  =  1,  is  the  point  A,  6  spaces  to  the  right  and 
1  space  up.  The  second  point,  x  =  5,  y  =  0,  is  the  point  B,  5  spaces  to 
the  right,  on  the  x-axis.  The  third  point,  at  =  4,  jf  =— 1,  is  the  point  C, 
4  spaces  to  the  right  and  1  space  down,  and  so  on.  Points  represented  by 
fractional  values  of  x  and  y  are  located  in  the  same  way.  Thus  x  =  1  £, 
y  —  3§,  locate  the  point  F. 

The  values  of  x  and  y  which  satisfy  the  equation  x  —  y  =  5 
are  represented  by  points  which  form  a  straight  line.  Further- 
more, every  point  in  the  line  corresponds  to  values  of  x  and  of  y 
which  satisfy  the  equation. 

Thus  the  graph  of  a  linear  equation  in  two  unknowns  is  a 
straight  line. 


EXERCISES 

91.  Tabulate  five  pairs  of  values  of  x  and  y  which  satisfy 
each  of  the  following  equations.  Draw  a  separate  pair  of 
axes  for  each  equation  and  plot  the  points. 

1.  2x  +  3y  =  6.  3.   x-y  =  2.  5.   x  =  2y. 

2.  Sx  —  4y  =  12.  4.   x  +  y  =  0.  6.   y  =  2x. 

92.  In  the  preceding  exercises  the  student  has  observed  that 
a  linear  equation  in  two  unknowns  produces  a  graph  which  is 
a  straight  line.  As  two  points  determine  the  position  of  any 
straight  line,  it  is  necessary  and  sufficient  to  find  only  two  points 
on  the  graph  of  a  linear  equation  in  order  to  fix  the  position  of 
the  line.  The  two  points  most  easily  found  are  x  =  0,  y  =  ? 
and  x  =  ?,  y  =  0. 

If  x  =  0  and  y  =  0  satisfy  the  equation,  it  will  be  necessary 
to  let  x  have  some  value  other  than  0  and  to  find  the  cor- 
responding value  of  y. 


70 


ELEMENTARY   ALGEBRA 


For  instance,  in  2a;  —  5  y  =  0,  if  a;  =  0,  y  =  0,  and  if  y  =  0,  x  =  0. 

But  if  x  =  5,  y  =  2. 

Plot  (0,  0),  and  (5,  2),  and  draw  the  line,  as  in  Fig.  14. 


Y 

o 

X 

H 

'0 

-V 

' 

EXERCISES 

93. 

Represent  the  fol- 

lowing 

equations   graph- 

ically: 

1. 

ic +*/=  3. 

2. 

x  —  y  =  5. 

3. 

2x  +  y  =  A. 

4. 

x  —  2y  =  6. 

5. 

2  a>  -  4  y  =  10. 

6. 

x  +  2y  =  0. 

7. 

a;  — 3y  =  0. 

8. 

3x-y  =  0. 

9. 

5x  =  <iy. 

Fig.  14 
PRACTICAL  APPLICATIONS   OF  GRAPHS » 

94.  One  of  the  simplest  applications  of  graphs  is  in  the  re- 
duction of  denominate  numbers  from  one  unit  to  another  with- 
out the  labor  of  numerical  computation.  Thus,  Fig.  15  enables 
us  by  inspection  to  reduce  miles  to  Kilometers,  or  Kilometers 
to  miles.  The  following  example  shows  how  the  graph  may  be 
drawn. 

95.  If  10  miles  are  equal  to  16.1  Kilometers,  how  many  Kilo- 
meters are  equal  to  x  miles  ? 

Let  y  be  the  required  number  of  Kilometers. 

x  mi.         y  Km. 
10  mi._  16.1  Km.* 


Then  by  proportion, 
Omitting  the  names  of  units, 


Multiply  both  sides  by  16.1, 
Or 


x  _     y 
i0~16l' 
1.61  x  =  y. 
y  =  1.61  x. 

1  May  be  postponed  for  the  present. 


PROPORTION 


71 


> 

1 

.210 
S 
B 

i 

^ 

C 

1 

l 

1 

1 

i 

10        D 


20 

Kilometers 

Fig.  15 


30 


The  graph  of  this  equation  is  a  straight  line. 
To  draw  this  graph,  carefully  locate  two  points  on  it  and 
then  draw  a  straight  line  through  the  two  points.     We  obtain 

y  =  1.61a; 


X 

y 

Point 

0 
10 

0 
16.1 

O 
A 

In  the  graph,  Fig.  15,  Kilometers  are  measured  along  the 
x-axis,  miles  are  measured  along  the  y-axis. 

To  determine  how  many  Kilometers  are  equal  to  8  miles,  we  find  the 
point  B  on  the  «/-axis  which  indicates  8  miles,  then  proceed  to  the  right 
(in  a  direction  parallel  to  the  x-axis)  to  the  point  C  on  the  graph,  then 
proceed  downward  to  the  point  D  on  the  x-axis.  The  number  of  Kilo- 
meters equivalent  to  8  miles  is  seen  to  be  approximately  13. 

Results  that  are  absolutely  accurate  cannot  be  obtained  for  three 
reasons :  First,  the  relation  "  10  miles  =  16.1  Kilometers"  is  correct  only 


72  ELEMENTARY   ALGEBRA 

to  tenths  of  Kilometers  ;  secondly,  even  if  this  relation  were  accurate,  the 
graph  could  not  be  drawn  with  absolute  accuracy  ;  third,  to  estimate  the 
fraction  of  a  Kilometer  at  the  point  C,  without  some  slight  error,  is  hardly 
possible.     Of  necessity  all  measurements  are  only  approximate. 

ORAL   EXERCISES 

96.  Estimate,  to  the  first  decimal,  the  number  of  Kilometers  in 

1.  7  mi.  4.    13  mi.  7.    17  mi. 

2.  9  mi.  5.    14  mi.  8.    18  mi. 

3.  11  mi.  6.    16  mi.  9.   20  mi. 

Estimate,  to  the  first  decimal,  the  number  of  miles  in 

10.  8    Km.  13.   17  Km.  16.    23  Km. 

11.  12  Km.  14.    19  Km.  17.    27  Km. 

12.  13  Km.  15.    21  Km.  18.   32  Km. 

CONSTRUCTION   AND  USE  OF  GRAPHS 

97.  1.  10  Kilograms  are  approximately  equal  to  22  pounds. 
Draw  a  graph  for  reducing  at  sight,  Kilograms  to  pounds,  and 
pounds  to  Kilograms.  Change  at  sight,  6  lb.,  7  lb.,  9  lb.  to 
Kilograms;  change  also  21  Kg.,  17  Kg.,  13  Kg.  to  pounds. 

2.  5\  yards  =  1  rod.  Construct  a  graph  for  converting 
yards  to  rods  and  rods  to  yards.  At  sight  change  to  yards, 
2.2  rd.,  1.7  rd.,  1.3  id.,  .8  rd. 

Let  1  in.  along  the  x-axis  stand  for  1  rd.,  and  \  in.  along  the  y-axis 
stand  for  1  yd.  This  gives  a  graph  which  is  convenient  for  reducing  a 
small  number  of  rods  to  yards. 

3.  Draw  a  graph  for  finding  the  lengths  of  circles  when 
the  diameters  are  given,  and  vice  versa. 

Take  the  length  of  a  circle  equal  to  ty  times  its  diameter. 

4.  Draw  a  graph  for  converting  temperatures  on  the 
Fahrenheit  scale  to  temperatures  on  the  Centigrade  scale,  and 
vice  versa. 

We  know  that  a  temperature  of  0°C.  is  the  same  as  one  of  32°  F. ; 
also  that  a  temperature  of  20°  C.  is  the  same  as  one  of  68°  F.    From  these 


PROPORTION 


73 


data  we  locate  the  points  A  and  B,  in  Fig.  16,  which  determine  the  line. 
This  graph  is  applicable  to  temperatures  below  zero,  or  negative  tem- 
peratures. 


Fig.  16 


5.  In  Fig.  16  change  the  following  to  temperatures  on  the 
Fahrenheit  scale :  30°  C,  25°  C,  17°  C,  0°  C,  -  10°  C,  -  20°  C. 

6.  In  Fig.  16  change  the  following  to  temperatures  on  the 
Centigrade  scale:  80° F.,  65° F.,  52° F.,  32° F.,  15° F.,  8°F., 
0°F.,  -3°F. 

7.  A  wholesale  dealer's  profit  when  he  sells  to  a  retail 
merchant  is  20%.  Draw  a  graph  for  ascertaining  the  cost 
price  when  the  selling  price  is  known,  and  vice  versa. 

When  there  is  a  profit  of  20%,  the  selling  price  is  to  the  cost,  as  120 
is  to  100. 

Let  x  be  the  selling  price  and  y  the  corresponding  cost. 


Then,  by  proportion, 


x   _   y 
120      100 ' 


Multiply  both  sides  by  100,  f  *  =  y. 

Or  y  =  %'*■■ 

The  graph  of  this  equation  is  a  straight  line. 


74  ELEMENTARY   ALGEBRA 

From  this  equation  the  following  data  are  obtained 


X 

y 

Point 

0 
30 

0 

25 

0 
A 

Through  the  points  O  and  A,  in  Fig.  17,  draw  a  straight  line,  which  is 
the  graph  required.  It  shows  the  relation  between  the  cost  and  the  sell- 
ing price. 

To  determine  the  cost  of 
an  article  that  sells  for  §  36, 
take  the  point  B  36  divisions 
from  O  on  the  x-axis,  then 
OB  represents  the  selling 
price.  From  B  pass  verti- 
cally to  C  on  the  "  wholesale 
to  retail"  graph,  thence 
horizontally  to  D.  Then 
OD,  or  §30,  is  the  cost  to 
the  wholesale  dealer. 


$10 


B$40 


$20         $30 
Selling  Price 

Fig.  17  8.    In    Fig.    17,    find 

the  cost  of  articles  sold 
to  the  retail  dealer  for  $10,  $14,  $17,  $24,  $28,  $32. 

9.    In  Fig.  17,  find  the  selling  price  to  the  retail  dealer  of 
articles  which  cost  $5,  $15,  $20,  $23,  $25. 

10.  A  wholesale  dealer  makes  a  profit  of  15  %  when  he  sells 
to  a  retail  dealer.  The  retail  dealer  makes  a  profit  of  65  °J0 
when  he  sells  to  a  consumer.  Draw  a  graph  for  finding  at 
sight  the  cost  to  the  wholesale  dealer  of  articles  whose  selling 
price  to  the  consumer  is  known. 

When  there  is  a  profit  of  15%,  the  selling  price  is  to  the  cost  as  115  is 
to  100. 

Let  x  be  the  selling  price  to  the  retail  dealer  and  y  the  cost  to  the 
wholesale  dealer. 

Then,  by  proportion,     -2-  =  JL. 
^         '     115     100 


PROPORTION 


75 


Multiply  both  sides  of  the  equation  by  100,  \^  x  =  y. 

When  x  =  0,    y  =  0,    hence  the  graph  passes  through  O. 

When  x  =  46,  y  =  40,  hence  the  graph  passes  through  A. 

Draw  the  line  OA. 

The  retail  dealer  sells  at  a  profit  of  65  %.  That  is,  his  selling  price  to 
the  consumer  is  to  his  buying  price  as  165  is  to  100. 

Let  x  be  the  selling  price  to  the  consumer  and  y  the  cost  to  the  retail 
dealer. 

Then,  _*_=JL. 

165      100 

Or  y  =  ftx. 


When  x  =  0,    y  =  0,    hence  the  graph  passes  through  O  in  Fig.  18. 
When  x  =  66,  y  —  40,  hence  the  graph  passes  through  B. 
Draw  the  line  OB. 


$10       $20 


$30 


$40         $50 
Selling  Price 

Fig.  18 


$60        $70        $80 


To  determine  the  cost  to  the  wholesale  dealer  of  an  article  for  which 
the  consumer  pays  .$60,  two  steps  are  necessary. 

The  first  step  is  to  find  its  cost  to  the  retail  dealer.  Take  the  point  C 
60  divisions  on  the  x-axis,  then  OC  represents  the  price  paid  by  the  con- 
sumer.    From  C  pass  vertically  to  D  on  the  "retail  to  consumers"  graph, 


76  ELEMENTARY   ALGEBRA 

thence  horizontally  to  E.    Then  OE,  or  about  $52.20,  is  the  cost  to  the 
retail  dealer. 

The  second  step  is  to  find  the  cost  to  the  wholesale  dealer.  Take  the 
distance  OF  equal  to  OE ;  from  F  pass  vertically  to  G  on  the  "  wholesale 
to  retail "  graph,  thence  horizontally  to  H.  Clearly  OH  is  the  cost  to  the 
wholesale  dealer,  about  $31  £.    This  is  approximately  the  required  answer. 

11.  What  is  the  cost  to  the  wholesale  dealer  of  articles 
which  are  sold  to  the  consumer  for  $  20  ?    $  30  ?    $  50  ?    $  75  ? 

12.  What  does  the  consumer  pay  for  articles  which  cost  to 
the  wholesale  dealer  $  50  ?    $  40  ?    $  15  ? 


\ 


-1      /rL 


L-    % 


2/ 


LJ 


'(X    —  - 


(X 


V.^ 


x*6 


^  I 


"2- 


0«* 


«y 


CHAPTER   V 

EQUATIONS  INVOLVING  FRACTIONAL   COEFFICIENTS 


98.    Solve  |  x  +  \  x  =  15.  ,      Xi 


First  Solution.   .  i  +  I  =  f  •  "*■        f- 

Hence  we  have  f  £  =  15. 

15   6 
Divide  both  sides  by  f,  x  =  — —  —  18.  Ans. 

5 

Second  Solution.  $  x  +  I  x  =  16. 

Multiply  both  sides  by  6,  2x  +  3  x  —  90. 

5x  =  90. 
x  =  18.  Ans. 
Check:  ^  (18)  +  i  (18)  =  15. 

6  +  9  =  15. 
15=  15. 
In  the  second  solution  we  multiply  both  sides  of  the  equation  by  some 
number  which  will  remove  the  denominators.     It  is   usually  easiest  to 
select  the  least  number  into  which  each  denominator  will  go  without  a 
remainder.     In  the  example  above,  that  number  is  6. 

EXERCISES 
99.    Solve  and  check : 
,   £.»  ,   Z^  =  6_r. 

2.    3y  +  ly  =  l.  g     2y  +  3      y-3=o. 

2jy  =  5_2v  5  3 

•     3       6      3'  9    y-1     y-2  =  2     y-3, 

.     n  ,   0         5n      3n  ,  1Q  2  3         3        4 

4.  -  +  2n = h  13. 

8  6        4  10    tn=Z.t  +  B     t  +  2=i 

5.  2,  +  5*.*£±2+».    X  ^  3  6 

^7  4  u.   |(6-x)  =  5. 

12     4  3  11         5V         < 

77 


78  ELEMENTARY   ALGEBRA 

13.  $(«  +  l)=2.  16.    .2*  =  48 -.04a;. 

14.  i(s  +  4)-is-8  =  0.  17.    .2w  + 3  =  3.8+ .04 w. 

15.  .5cc  =  3. 

18.  $(*  +  2)  +  f(*-12)  =  |(*  +  9). 

19.  ^(„_l)_i(„_2)  =  f-4>-3). 

20.  i(2/  +  l)-|(y  +  4)  +  i(2/  +  3)  =  16. 

PROBLEMS 

100.  1.  One  half  a  certain  number  plus  one  fourth  that 
number  increased  by  one  third  the  number  equals  26.  Find 
the  number. 

2.  The  sum  of  two  numbers  is  29.  One  half  the  first  plus 
one  third  the  second  is  12.     What  are  the  numbers  ? 

3.  The  difference  between  two  numbers  is  6.  One  ninth 
of  the  first  plus  one  third  of  the  second  is  6.  Find  the 
numbers. 

4.  Separate  48  into  two  parts  so  that  ^  the  larger  minus  ^ 
the  smaller  equals  £  the  number  itself. 

5.  One  fourth  of  a  certain  number  increased  by  1^,  and  the 
result  diminished  by  the  quotient  obtained  by  dividing  the  sum 
of  twice  the  number  and  4  by  9,  equals  1.     Find  the  number. 

6.  The  width  of  a  rectangle  is  4  of  its  length  and  its  pe- 
rimeter is  54  in.     Find  its  dimensions. 

7.  The  length  of  a  rectangle  is  12  in.  more  than  the  width. 
The  sum  of  the  length  and  width  is  twice  their  difference. 
Find  the  area. 

8.  The  width  of  a  rectangle  is  5  ft.  less  than  the  length. 
If  the  length  be  decreased  2  ft.  and  the  width  increased  3  ft., 
the  area  will  be  increased  11  sq.  ft.     Find  the  area. 

9.  A  certain  square  has  the  same  area  as  a  rectangle  whose 
dimensions  are  3  ft.  longer  and  2  ft.  shorter  than  those  of  the 
square.     What  is  the  area  of  each  ? 


FRACTIONAL   COEFFICIENTS  79 

10.  Find  two  consecutive  integers  such  that  £  the  first  plus 
\  the  second  equals  7  less  than  the  first. 

11.  Find  three  consecutive  even  integers  such  that  \  of  the 
first  and  \  of  the  second,  plus  \  of  the  third,  equals  the  third. 

12.  What  three  consecutive  integers  have  their  sum  equal 
to  48  ?  How  does  the  sum  compare  with  the  second  integer  ? 
Is  this  true  of  any  three  consecutive  integers  ? 

13.  The  sum  of  |  of  a  certain  even  integer,  \  of  the  next 
odd  one,  and  %  of  the  next  even  one  equals  the  even  integer 
just  before  the  one  first  mentioned.  Find  the  series  of  in- 
tegers. 

14.  A  is  15  years  old ;  B  is  25  years  old.  In  how  many 
years  will  A  be  f  as  old  as  B  ? 

15.  A  is  twice  as  old  as  B ;  five  years  ago  he  was  2\  times 
as  old.     How  old  is  each  now  ? 

16.  A  man's  age  6  years  ago  was  \  of  what  his  age  will  be 
30  years  hence.     Find  his  age  now. 

17.  A  father  is  3  times  as  old  as  his  son ;  in  5  years  the 
father  will  be  2\  times  as  old  as  the  son.  What  is  the  age  of 
each? 

18.  A  man  who  is  35  years  old  has  a  son  10  years  old.  In 
how  many  years  will  the  boy  be  \  as  old  as  his  father  ? 

19.  A  man  gave  to  one  son  \  of  his  money,  to  a  second  \  of 
his  money.  He  had  left  $  10  less  than  the  number  of  dollars 
he  gave  to  both.     How  much  had  he  at  first  ? 

20.  I  spent  $1.75  for  \$  and  2f  stamps,  buying  25  more  of 
the  former  than  of  the  latter.  How  many  of  each  kin^l  did  I 
buy? 

21.  I  took  a  trip  of  105  miles,  partly  by  trolley,  p*artly  by 
train.  If  I  went  \  as  far  again  by  train  as  by  trolley,  how 
far  did  I  go  by  each  ? 


80  ELEMENTARY   ALGEBRA 

22.  A  collection  of  5^  pieces  and  quarters  amounts  to  $  1.60. 
There  are  2  more  5^  pieces  than  quarters.  How  many  are 
there  of  each  ? 

23.  I  have  three  more  half  dollars  than  quarters.  The 
value  of  the  half  dollars  exceeds  the  value  of  the  quarters  by 
$4.50.     How  much  money  have  I  ? 

24.  Thirty  coins  —  dimes,  nickels,  and  quarters  —  amount 
to  $5.30.  There  are  i  as  many  nickels  as  quarters.  How 
many  are  there  of  each  ? 

25.  Three  boys  have  150  marbles.  If  the  third  has  f  as 
many  as  the  second,  and  the  second  £  as  many  as  the  first, 
how  many  marbles  has  each  boy  ? 

26.  The  same  number  is  subtracted  from  60  and  from  45. 
One  third  of  the  first  remainder  equals  \  of  the  second.  What 
is  the  number  ? 

27.  In  going  a  certain  distance,  a  train  traveling  at  the  rate 
of  40  miles  an  hour  takes  21  hours  longer  than  one  traveling 
50  miles  an  hour.     What  is  the  distance  ? 

Use  d  =  rt. 

28.  I  have  13  hours  at  my  disposal.  How  far  may  I  go  at 
the  rate  of  3  miles  an  hour,  so  as  to  return  home  in  time,  corn- 
ing back  at  the  rate  of  3^-  miles  an  hour  ? 

29.  A  man  sold  a  home  for  $4944,  gaining  3%  of  the  cost. 
How  much  did  the  home  cost  ? 

Hint,     x  +  .03  x  =  4944. 

30.  An  agent  deducted  5  %  commission  for  the  sale  of 
property,  remitting  $  6222.50.     Find  the  selling  price. 

31.  A^  6%  interest  for  a  certain  time,  $350  amounted  to 
$444.50.     What  was  the  length  of  time  ? 

Use  t"  =  prt. 

32.  The  interest  of  $365  for  2  years  3  months  at  a  certain 
rate  was  $  32.85.     What  was  the  rate  ? 


FRACTIONAL  COEFFICIENTS  81 

33.  The  amount  of  a  certain  principal  at  6  %  in  5^  years  is 
$  7714.     What  is  the  principal  ? 

34.  A  man  invests  a  part  of  $  1500  at  5  %  and  the  re- 
mainder at  4  <f0.  The  5  °J0  investment  yields  $39  more  a  year 
than  the  4  %.     How  much  has  he  invested  at  4  <f0  ? 

35.  A  part  of  $  3680  is  invested  at  5|  °f0  and  the  remainder 
at  6%.  The  total  interest  in  2  years  amounted  to  $411.60. 
How  many  dollars  are  there  in  each  investment  ? 

36.  If  the  valuation  of  a  certain  property  is   $  17,000,  and 

the  annual  tax  thereon  is  $161.50,  find  the  tax  rate  in  mills 

on  a  dollar. 

Let  x  mills  =  the  tax  on  a  dollar.    Then  17000a;  =  161.50. 

1000 

Find  the  tax  rate  in  mills  on  a  dollar : 

Valuation  Tax  Valuation  Tax 

37.  $6,560  $55.76         40.    $125,500         $909,875 

38.  $29,800         $186.25         41.    $375,400         $2909.35 

39.  $90,700         $589.55         42.    $875,500         $6566.25 

43.  I  wish  to  get  $  7280  from  a  bank.  For  what  sum  must 
I  make  out  a  60-day  note,  to  obtain  this  sum,  bank  discount 
being  at  the  rate  of  6  °/0  per  annum  ? 

Bank  discount  is  computed  on  the  face  of  the  note. 

Let  8  x  s  the  required  sum.  The  bank  discount  for  60  da.  is  1  %  of 
|  x,  or  8  .01  x.  Subtracting  8  .01  x  from  $  x,  leaves  8  .99  x,  the  proceeds. 
Hence,  .99  x  =  7280. 

44.  A  man  needs  $  2975.  For  what  sum  must  he  make  out 
a  60-day  note  to  obtain  the  $  2975  as  proceeds,  if  the  bank  dis- 
counts the  note  at  6  %  per  annum  ? 

45.  Mr.  Murray  makes  out  a  90-day  note,  which  is  dis- 
counted by  the  bank  at  2  °f0  for  the  90  days.  What  must  be 
the  face  of  the  note,  to  yield  $  9700  as  proceeds  ? 

46.  John  Allen  issued  a  note  to  a  bank  which  secured  for 
him,  after  it  was  discounted  at  3  %  for  the  term  of  4  months, 
the  sum  of  $  29,750.     Find  the  face  of  the  note. 


82  ELEMENTARY   ALGEBRA 

The  federal  income  tax  in  the  United  States  requires  that  every  unmar- 
ried person  shall  pay  an  annual  tax  of  1  %  on  his  net  income  in  excess  of 
$  3000,  and  that  married  persons  living  together  shall  pay  an  annual  tax  of 
1  %  on  their  joint  income  in  excess  of  $4000.  When  the  net  income  ex- 
ceeds $  20,000,  an  additional  tax  thereon  is  levied  as  follows  : 

1  %  on  part  of  income  over  §  20,000  and  not  above  $  50,000. 

2%  on  part  of  income  over  f  50,000  and  not  above  $  75,000,  etc. 

Thus,  an  unmarried  man  pays  on  an  income  of  $  70,000  the  following 
tax  :  1  %  on  $  67,000  +  1  %  on  9  30,000  +  2  %  on  $  20,000. 

47.  An  unmarried  person  whose  net  annual  income  is  be- 
low $20,000  pays  an  income  tax  of  $145.70.  Find  his  in- 
come. 

48.  A  man  and  wife  have  a  joint  income  that  is  below 
$  20,000 ;  their  income  tax  is  $  135.45.     Find  their  income. 

49.  A  man  and  wife  pay  an  income  tax  of  $200.  Does 
their  annual  income  exceed  $  20,000  ?     Find  their  income. 

Let  $  x  =  income.     Then  .01(x  -  4000)  +  .01(x  -  20,000)  =  200. 

50.  The  annual  income  tax  of  an  unmarried  man  is  $450. 
Is  his  income  above  $20,000?     Above  $50,000?     Find  it. 

51.  The  joint  income  tax  of  a  man  and  his  wife  is  $  1360.  Is 
the  annual  income  above  $50,000?  Above  $75,000?  Com- 
pute their  income. 

52.  What  is  the  annual  net  income  of  an  unmarried  man 
whose  income  tax  is  $  850  ? 

53.  A  boy  on  a  farm  receives  2\$  a  dozen  for  delivering  98 
dozen  eggs  and  agrees  to  have  27^  deducted  from  his  pay  for 
every  dozen  eggs  that  he  breaks  and  does  not  deliver.  He  is 
paid  $  1.55.     How  many  dozen  eggs  did  he  break  ? 

54.  Two  persons  traveling  toward  each  other  set  out  at  the 
same  time  from  towns  198  miles  apart.  One  person  walks  10 
miles  a  day,  the  other  12  miles.  In  what  time  will  they  meet, 
and  how  many  miles  will  each  have  walked  ? 

55.  At  an  election  \  of  a  constituency  abstained  from  voting, 
and  |  of  the  constituency  voted  for  the  successful  candidate. 


FRACTIONAL   COEFFICIENTS  83 

At  a  subsequent  election  f  abstained  from  voting,  and  i  voted 
for  the  successful  candidate.  The  minority  in  the  latter  elec- 
tion was  120  less  than  in  the  former.  Find  the  number  of 
voters  in  the  constituency. 

56.  A  rectangular  tank  is  10  ft.  6  in.  long  and  7  ft.  3  in. 
wide,  and  contains  400  gallons  of  water.  Find  the  depth  of 
the  water,  correct  to  one  tenth  of  an  inch,  having  given  that  a 
gallon  contains  231  cu.  in. 

57.  A  train  is  5  minutes  late  when  it  performs  its  journey 
at  the  rate  of  29^  miles  an  hour,  and  is  2  minutes  late  when  it 
travels  at  30  miles  an  hour.  What  is  the  number  of  miles  in 
the  journey  ? 

58.  There  are  two  lawns  in  one  garden.  One  is  square ;  the 
second  is  3  yd.  narrower  and  4  yd.  longer  than  the  first,  but 
of  equal  area.     Find  the  area  of  each. 


CHAPTER   VI 

SPECIAL  PRODUCTS 

101.   I.   (a  +  by.  ■ 


a  +  b 

a  +  b 

a?  +  ab 

+  ab  +  b2 

a2  +  2  ab  +  62 

II.    (a -by. 

a  —  b  . 

a  —  b 

■ 

a2—  ab 

-ab  +  b2 

a2 -2  ab  +  b2 
A  (a  4-  &)2  =  a2  +  2  a&  +  ft2. 
(a  -  d)2  =  a2  -  2  a&  +  ft2. 

\  of  two  numbers  is  equal 
difference  J 

to  the  square  of  the  first,  \         ■    \  twice  the  product  of  the  first 

{minus} 

and  the  second,  plus  the  square  of  the  second. 

ORAL   EXERCISES 

102.   Find,  by  inspection,  the  values  of : 

1.  (m  +  w)2.  5.    (4  + a)2. 

2.  (c-d)2.  6.    {b-iy. 

3.  (a-2  6)2.  7.    (x2-l)2. 

4.  (3x  +  yy.  8.    (2a2-3  6)2. 

84 


SPECIAL   PRODUCTS 


85 


9.  (4c  +  3eF)2. 

10.  (7ab-2y. 

11.  (9x2-2y2)2. 

12.  (oa3  +  4x2)2. 

13.  (1+z)2. 

14.  (|a-3)2. 

15.  (!c-icf>'. 


i6.  (cc°  +  y-)2. 

17.  (3am-2  6»)2. 

18.  (7a-6)l 

19.  (5d-3e)2. 

20.  (ra  +  12n)2. 

21.  (9  a  —  3  y)2. 

22.  (2 a- 7  bf. 


23.  What  must  be  added  to  a2  +  10  a  to  make  it  the  square 
of  a  +  5  ? 

24.  What  must  be  added  to  x2  —  xy  +  y2  to  make  it  the 
square  of  x  —  y  ? 

25.  What  must  be  added  to  36  —  12  a2  to  make  it  a  perfect 
square  ? 

In  the  following,  supply  terms   necessary  to  form  perfect 
squares : 


26.  c2+(  )+d2. 

27.  m2+(  )  +  4n2. 

28.  x2+(  )+9. 

29.  t2-(  )+16. 

30.  4a2-(  )+l. 

31.  9x2+(  )+36. 

32.  16  62-(  )+49. 

33.  25a4-10a2+(  )• 


34.  6±y2  +  32y  +(  ). 

35.  121m6-22m3+(  ). 

36.  9a10  +  12a562  +  (  ). 

37.  (  )-6&2  +  9. 

38.  (  )+10x4y3  +  25y6. 

39.  (  )+80M2  +  64fc4. 

40.  (  )-24s5  +  144. 

41.  (  )-  182?n2+169. 


Find  the  binomials  whose  squares  are : 


42.  x2  +  6  x  +  9. 

43.  4  a2  —  4a&  +  &2. 

44.  x2  +  22  xy  + 121  y2. 

45.  1  -  14  y  +  49  y2. 

46.  m2  +  40  m  +  400. 

47.  n2  —  4n  +  4. 


48.  16*/2-8y  +  l. 

49.  25  c*  +  160^  +  256  c2. 

50.  9  a2  -  30  aa^2  +  25  xy . 

51.  36  62  +  60  a^c3  +  25  a*c*. 

52.  64  c8  -  16  c4d5  +  d10. 

53.  4  a%2  +  28  xy  +  49. 


86 


ELEMENTARY   ALGEBRA 


103.   III.   (a  +  b)(a-b). 

a  +  b 
a-b 


a2  +  ab 

-ab-b2 
a2  -6« 

.-.  (a  +  6)(a-6)  =  a2-62. 

Rule.  Tlie  product  of  the  sum  of  two  numbers  and  their  differ- 
ence is  equal  to  the  square  of  the  first  number  minus  the  square 
of  the  second. 


ORAL   EXERCISES 

104.    Find,  by  inspection,  the  values  of 

1.  (k  +  y)(k-y). 

2.  (l-3a)(l+3o). 

3.  (2h  +  7)(2h-7). 

4.  (5 -&)(&  + 5). 

5.  (10-3a)(3a  +  10). 

6.  (3n  +  7)(3»-7). 

7.  (3x  —  a)(3x  +  a). 

8.  (5a6  +  3)(5a6-3). 

9.  (2  n  +  7)(7  -  2  n). 
10.  (6&-l)(6  6  +  l). 


11.  (a2-9  62)(a2  +  9  62). 

12.  (LSa^  +  yXlSar'-y). 

13.  (4  aft  —  c)(4  oft  +  c). 

14.  [(o  +  6)+c][(o+6)-c]. 

15.  (aj  +  6  +  2)(x  +  6-2). 

16.  [(m— n)— 3][(m— n)+3]. 
[« +(S +  „)][«_  (S  +  W)].  ' 

[7+»+y][7-*-sfj. 

[5+(o-6)][5-(a-6)]. 
(2  +  c-d)(2-c  +  d). 


17 
18 
19 

20 


Find  two  binomials  whose  product  is  : 

21.  x*  —  4  f.  28.  121  c2d2  —  144  aj4. 

22.  16 -a4.  29.  225b* -1. 

23.  4x2-16.  30.  1-81  a10. 

24.  9a2-64  64.  31.  100  62  -  36. 

25.  25m4-49w6.  32.  ^  -  225. 

26.  a4  — 9.  33.  ff -a6 

27.  9a%2-25.  34.  16m2-25?i2. 


SPECIAL   PRODUCTS  87 

35.  (x  —  y)2  —  z2.  38.    a2  -  (b  +  c)2. 

36.  (c  +  d)2  -  e2.  39.    t2  -  (s  -  u)2. 

37.  (m-n)2- 64.  40.    81-(a-6)2. 

In  each  of  these  exercises  (21-40),  the  two  binomials  found  are  called 
factors  of  the  given  expression.  Thus,  in  Ex.  21,  x3  +  2  y3  and  x3  —  2  y3 
are  said  to  be  factors  of  x6  —  4  j/6. 

105.   IV.   (*+a)(as  +  &). 

a;  +  a 


a2  +  era; 

-4-  bx  +  q& 

#2-f(a4-  b)x+  ab 

.:  (x  +  a)(x  +  b)  =  x2  +'(a  +  b)x  +  ab. 

Rule.  The  product  of  tico  binomials  having  a  commpn  term 
is  equal  to  the  square  of  the  common  term,  and  the  sum  of  the  un-, 
like  terms  times  the  common  term,  plus  the  product  of  the  unlike 
terms. 

ORAL,   EXERCISES 

106.    Find,  by  inspection,  the  values  of : 

1.  (x  +  3)(a,-  +  4).  11.  (a  +  4c)(a  —  c). 

2.  (x  -  3)(x  -  4).  12.  (2t  +  3)(2t-5). 

3.  (*  +  3)(*  -  4).  13.  (a>  -  16)(»  +  20). 

4.  (s-3)(«  +  4).  14.  (3ft-l)(3ft-19). 

5.  (y  +  2)0/  -  7).  15.  (m  +  17)(m-2). 

6.  (a  — 9)(o  — 3).  16.  (c  —  x)(c  +  3x). 

7.  (6-5)(6  +  6).  17.  (s-8)(s-3). 

8.  (ax  +  l)(ax-2).  18.  (7h-15)(7h  +  2). 

9.  (v-o)(»-2a).  19.  (4gr-a2)(4gr  +  5a2). 
10.  (n  +  b)(n  +  46).  20.  (5  c2  -  9)  (5  c2  -  4). 


88 


ELEMENTARY   ALGEBRA 


Find  two  binomials  (two  factors)  whose  product  is 


21.  a2  +  7  a  +  12. 

22.  a2-9a  +  20. 

23.  62  +  26-35. 

24.  a2 +  11  a -12. 

25.  c2- 20  c +  51. 

26.  m2-  25  m  +  100. 

27.  *2-25«  +  66. 

28.  4r2  +  22r  +  30. 

29.  25  s2 -30  s +  8. 

30.  9eP-60d  +  19. 


31.  100  m2  +  10  m  -  12. 

32.  x2y2  +  12xy-  64. 

33.  16s4  +  16s2-5. 

34.  15  —  8  m  +  ra2. 

35.  8 -42  a +  49  a2. 

36.  m2- 33  m +  90. 

37.  121 1*  -  77  t  -  18. 

38.  962-9  6-40. 

39.  100  z2  -  40  a  -  12. 

40.  /2-13/-30. 


CHAPTER   VII 

FACTORING 

107.  In  the  preceding  chapter  we  found  the  products  of 
certain  binomials.  We  performed  also  the  inverse  operation  : 
given  the  product,  to  find  the  binomials. 

For  instance,  starting  with  the  product  x2  —  4  y2,  we  found  two  bi- 
nomials, x  +  2  y  and  x  —  2  y,  whose  product  is  x1  —  4  y2. 

This  inverse  operation  is  called  factoring ;  x  +  2y  and  x  —  2 y  are 
called  factors  of  x2  -  4  y'2. 

In  this  chapter  we  enter  upon  a  fuller  treatment  of  factoring. 

Factoring  an  expression  is  the  process  of  finding  two  or  more 
expressions  which,  multiplied  together,  produce  the  given  ex- 
pression. 

In  factoring  it  is  frequently  necessary  to  find  roots  of  mono- 
mials. 

108.  The  square  root  of  a  monomial  is  one  of  the  two  equal 
factors  whose  product  is  the  monomial. 

Since  5  •  6  =  +  25,  and  (  —  5)  ( —  5)  =  -f  25,  it  follows  that  there  are  two 
square  roots  of  25  ;  namely,  4-  5  and  —  5. 

Similarly  there  are  two  square  roots  of  a2  ;  namely,  4-  a  and  —  a. 

The  two  square  roots  of  25  are  usually  written  in  the  form  ±  5  ;  sim- 
ilarly the  two  square  roots  of  a2  are  written  ±  a. 

The  positive  square  root  is  called  the  principal  square  root. 

We  write  V  25  =  5.  When  V  has  no  sign  before  it,  or  has 
the  +  sign  before  it,  the  principal  root  is  always  understood. 
When  we  write  —  V      we  mean  the  negative  root. 

Thus,  V25  stands  for  +  5,  V25  does  not  stand  for  —  5. 
—  V25  stands  for  —  5,  —  V25  does  not  stand  for  +  5. 
±  V25  =  ±  5. 

89 


90  ELEMENTARY   ALGEBRA 

Since  x?  •  x3  =  a6,  and  (—  &)(—  a^)=  jc6,  we  have  Vo6  =  +  a?, 
—  V&6  =  —  a,*3,  ±  Var6  =  ±  as3. 

Similarly,  Vl6a?  =  +  4x3,  -  V16  z6  =  -  4  x3,  ±Vl6a*  = 
±4  a?. 


Also,  ±  V4  a6b*  =  ±  2  a3b2,  ±  V36  a8610  =  ±  6  a465. 
How  is  the  exponent  of  a  letter  in  the  square  root  found  ? 

Rule.  The  two  square  roots  of  a  monomial  are  found  by 
writing  ±  the  square  root  of  the  numerical  coefficient,  times  the 
letters  of  the  monomial,  each  with  an  exponent  that  is  half  of  its 
exponent  in  the  given  monomial. 

Similar  results  hold  for  the  fourth  or  sixth  roots,  or  higher 
even  roots  of  monomials. 

Thus,  \/l6=  +  2,  -y/W=-2,  ±v^l6=±2,  V^STa*  =  +  3  a;2, 
-\/81a*=-3a;2,   ±  ^81  x8  =  ±  3  a;2,    ±vV=±a:,    ±v¥o«  =  ±2o2. 

109.  The  cube  root  of  a  monomial  is  one  of  the  three  equal 
factors  whose  product  is  the  monomial. 

Thus,  \/8  =  2,  because  2.2-2  =  8. 

Notice  that  v^T  is  not  -  2,  for  (-  2)(—  2)(—  2)  is  not  8. 

v^8=-2,  because  (-2)(- 2)(-2)  =  - 8. 

Observe  that  a  negative  number  can  be  the  cube  root  of  a  negative 
number,  but  a  negative  number  cannot  be  the  square  root  of  a  negative 
number. 

That  is,  -4  is  \/— 64,  because  (- 4)(— 4)(- 4)  =  - 64,  but  —  8  is 
not  V—  64,  because  (—  8)(—  8)  is  not  —  64. 

Nor  indeed  can  V—  64  be  +  8,  because  8  •  8  is  not  —  64. 

In  other  words,  neither  a  positive  number  nor  a  negative  number  can 
be  the  square  root  of  a  negative  number. 

From  V8  =  2  and  V  —  8  =  —  2  we  see  that  the  cube  root  of 
a  monomial  has  the  same  sign  as  the  monomial. 

Since  (3a2) (3 a2) (3 a2)  =  27  a6,  we  have  v/27o«"  =  3 a2. 

110.  The  square  root  of  2  cannot  be  exactly  expressed  by 
the  Hindu-Arabic  numerals.      One  can   only  approximate  its 


FACTORING 


91 


value- by  extracting  the  square  root  to  three  or  four,  or  more, 
decimal  places.  The  radical  y/2,  and  other  radicals  of  the 
same  kind,  like  V3,  Vo,  etc.,  whose  values  cannot  be  found 
exactly,  but  only  approximately,  represent  numbers  called 
irrational  numbers. 


ORAL  EXERCISES 
111.    Give  the  values  of: 

1.    VI2I. 
-V121. 
Jm2 


5.  ±  Vl6afy*. 

6.  -V25#*z«. 

7.  +V 100  aWc2. 


8.    Va^8*. 

Give  both  square  roots  of : 

13.  144  a12.  15.   225&hfhP 

14.  169  a668.  16.  100  r*s16. 


Give  the  values  of : 


19. 


125. 


20.  V-125. 

21.  ^16o46». 


23.  ^32. 

24.  -tfaWd* 
25 


22.    V64X6. 


V-216a9. 

26.  \/8^yv». 


9.    —  V|a2&2. 

10.  ±  V4  a^Y. 

11.  -V36~rW. 


12.    V49  m6n*oY- 

17.  400  a16^20. 

18.  900/rVr10. 


27.    VaMW2. 


28.  Va246«c72. 

29.  v^JV. 


30 


31.    Simplify  V32. 

32  is  equal  to  16  •  2,  where  16  is  a  perfect  square. 

Hence  V32  =  Vl6  •  2  =4\/2.  We  take  the  square  root  of  16,  which  is 
4,  and  write  the  4  as  a  factor  before  the  V  .  The  factor  2  is  not  a  perfect 
square  and  is  kept  under  the  radical  sign. 


32.    Simplify  Vo0a265. 

Here  50  c265  =  25  a2b*  times  2  b.    Since  25  a264  is  a  perfect  square,  we 
have  V50a265  =  5aft2v/26. 


92  ELEMENTARY   ALGEBRA 


Simplify  the 

following : 

5. 
6. 

7. 
8. 

1.  V200afy2. 

2.  V5l2. 

V12  aWcb. 
V300  m2np\ 

3.    Vl8a263. 

VSOOyzW. 

4.    V27aW. 

Vl28a62c3d4. 

112.  An  integral  number  or  expression  is  one  that  is  free 
from  fractions. 

For  example,  125  and  7  a  +  4  b2  are  integral  numbers  or  expressions. 

A  prime  factor  of  an  integral  number  is  an  integral  factor 
whieh  is  exactly  divisible  only  by  itself  and  one. 
Thus  3  and  7  are  prime  factors  of  63. 

A  prime  factor  of  an  integral  algebraic  expression  is  an 
integral  factor  which  cannot  itself  be  resolved  into  rational 
factors  other  than  itself  and  one. 

Since  (a2  +  2  b2){a2  —  2  b2)  =  a4  -  4  6*,  we  know  that  a2  +  2  b2  and 
rt2  —  2  b2  are  factors.  Moreover,  these  are  prime  factors.  By  multiply- 
ing we  see  that  (a  +  V2  b) (a  -  V2  6)  =  a2  -  2  ft2.  But  (a  4-  V2  6)  and 
(a  —  V2  6)  are  not  prime  factors,  because  they  contain  the  irrational  V2. 

In  this  chapter  we  shall  find  the  prime  factors  of  expres- 
sions ;  we  shall  not  search  for  factors  involving  irrationals. 

Factors  of  monomials  are  easily  found  by  inspection.  Thus 
the  monomial  30  a¥y  =  2*3-5  a?x*y  =  2  •  3  •  5  •  a  -a  •  x-x-x-y. 

If  a  polynomial  contains  a  monomial  factor,  it  may  be  found 
easily  by  inspection.  In  factoring  a  polynomial,  proceed  as 
follows  : 

(1)  Divide  the  polynomial  by  the  highest  monomial  factor 
common  to  all  the  terms. 

(2)  When  possible,  factor  the  quotient  thus  obtained. 

Thus,  in  32  a8b3  +  28  a2b2  —  12  ab,  we  see  that  each  term  contains  the 
factor  4  ab.  Divide  the  polynomial  by  4  ab  ;  the  quotient  is  8  a2b2  + 
7  ab  —  3,  which  (as  we  shall  see  later)  cannot  itself  be  resolved  into 
factors  free  from  inexact  roots. 

Therefore,  32  a3b8  +  28  a2b2  -  12  ab  =  4  ab(8  a2b2  +  7  ab  -  3). 


FACTORING  93 

113.  To  check  an  example  in  factoring  either 

(I)  Multiply  the  factors  together,  or 

(II)  Substitute  numbers  for  the  letters  and  simplify. 

In  82  aW  +  28  a262  -  12  ab  =  4  a&(8a262  +  7  ab  -  8),  let  a  =  b  =  1. 
Then  32  a363  +  28  flW  -  12  aft  =  48  ;  4  aft  =  4,  8  a262  +  7  aft  -  3  =  12, 
and  4  •  12  =  48.* 

114.  Since  factoring  is  the  inverse  of  multiplication,  certain 
type  forms  can  always  be  factored. 

I.    Type  form  :  ay  +  by  —  cy. 

ay  +  by  -  cy  =  y(a  +  b  -  c), 

wherein  y  stands  for  the   highest  monomial  factor  common 
to  all  the  terms. 

Hence  15  b3x  -  10  b3y  -  5  b3z  =  5  63(3  x  -  2  y  -  z). 

EXERCISES  » 


115 

.    Factor : 

l. 

4a-8. 

6. 

2  ra2^  —  6  m?n2  +  8  mhi3. 

2. 

X2  —  X3. 

7. 

15  a4  -  10  a3  -  25  a2. 

3. 

'Sab  —  6  ac. 

8. 

16  c3  +  8  c2d  -  64  c4^2. 

4. 

2  x2y  —  6  x*z. 

9. 

a;  —  ar2  +  x3  +  x4. 

5. 

a2b  +  a2c  —  a2d. 

10. 

12 a6  -24 a7  +  6a5. 

11.   x2  +  2 

xy- 

•  X  —  xz. 

12.  18d5-81(f-r-27d2-9d3. 

13.  17  ay  -  34  aV  +  68  a3. 

*  If  an  error  exists,  checking  by  the  substitution  of  particular  numbers  for 
the  letters  usually  reveals  the  error,  but  not  always.  There  is  the  possibility 
that  some  particular  numbers  will  be  hit  upon  which  fail  to  reveal  the  error. 
For  instance,  a2  —  62=£ (a2  —  b)(a  +  b)  (where  *fc  means  "  is  not  equal  to  "), 
yet  a  =  1,  b  =  2  yields  1  -  4  =  (1  -  2)  •  3,  or  -  3  =  -  3. 

Again,  a  =  2,  6  =  2  yields  0  =  2-0,  or  0  =  0.  Again  a  =  0,  6  =  10  yields 
—  100  =  —  100.  If  the  substitution  yields  an  inequality,  the  factoring  is  cer- 
tainly wrong  ;  if  the  substitution  yields  an  equality,  the  factoring  is  very 
probably  correct,  but  the  test  is  not  infallible  in  this  case. 


94  ELEMENTARY   ALGEBRA 

14.  3  xhjz  +  3  xyh  +  3  xyz2. 

15.  44  m?n2  —  55  m3n  +  66  m3r. 

16.  aW  -  aW  +  aW. 

17.  125  c2d2  +  100  <?d?  -  150  &<P. 

18.  15  6^/i2  +  17  bg2h2-byh  -  6  6g*. 

19.  a"6  —  aBc  —  and. 

20.  2a5°  — 6&x0  +  4cz0. 

21.  3  aB6"c  —  9  anbncd  +  6  anbnce. 

22.  cy2"  —  dy2n  -f  gyin. 

23.  5  a2&"  -+-  10  a?bn  — 15  a*bn. 

24.  8  cCy6  —  4  2/*  —  2  aftfz0. 

116.    II.    Type  form :  ax  +  ay  +  bx  +  by. 

Group  the  first  two  terms  together  and  the  last  two  terms 
together.  Then  factor  each  group.  Notice  that  the  factoring 
should  reveal  the  same  binomial  or  polynomial  factor  in  all 
the  groups. 

We  obtain 

(ax  +  ay)  +  (bx  +  by)=  a(x  +  y)  +  b(x  +  y). 

Divide  by  x  +  y  and  we  obtain  the  quotient  a  +  b. 
Taking  the  divisor  x  +  y  as  one  factor,  and  the  quotient  a  +  b  as  the 
other  factor,  we  have 

ax  +  ay  +  bx  +  by  =(z  +y)(a  +  b). 

Factor  2a-6b  — ad  +  3bd. 

Grouping,  (2  a  -  6  6)  -  (od  -  3  bd). 
Factoring  each  group,  2(a  —  3  6)  —  d(a  —  3  b). 
The  factor  a  —  3  b  is  seeu  to  occur  in  both  groups. 
Dividing  by  a  —  3  6,  we  obtain  2  —  d. 

Taking  the  divisor  a  —  3  6  as  one  factor,  and  the  quotient  2  —  d  as  an- 
other factor,  we  have 

2a-6b-ad+  3  bd  =(a  -3  6)(2  -  d). 

The  groups  may  contain  two  or  more  terms,  but  all  the  groups 
must  have  the  same  polynomial  factor.     In  grouping,  the  terms 


FACTORING  95 

may  be  taken  in  any  order  which  promises  to  reveal  a  common 
polynomial  factor. 

For  example,  ab  —  c  —  be  +  a  may  be  written  ab  —  be  +  a  —  c. 
Then  ab  -  6c  +  a  —  c  =  b(a  —  c)  -\-(a  —  c)  =  (a  —  c)(b  +  1). 

EXERCISES 

117.  Factor: 

1.  a(x  —  y)  +  b(x  —  y).  3.  x(c  -  a)  +  y(c  -  a). 

2.  a(x  +  2y)-b(x  +  2y).        4.  3(c  -  d)  -  z(c  -  d). 

5.  a(3  c  +  d)  -  6(3  c  +  d)  +  c(3  c  +  d). 

6.  —  2(a  +  #)  +  a(a  +  a)  —  e(a  4-  x). 

7.  b(a  —  c)  —  c(c  —  a). 

8.  r(«— y)+s(a>— y)+*(y— a>).     15.  y3  +  y2  +  y  +  1. 

9.  ax  —  ay  —  bx  +  by.  16.  a2—  a—  6  +  a&. 

10.  3  »  —  3  ?/  -  dx  +  dy.  17.  6  bx  -12  by-  50 cy+25 ex. 

11.  ac  +  ad  -  be  -  bd.  18.  3  b  —  5  by*  -  6  by  + 10  fry4. 

12.  xy  -  bx  -  ay  +  aft.  19.  49  ar  —  14  a<  +  21  &>•  —  6  6«. 

13.  cd  +  3c-2d  — 6.  20.  1+a  +  b  +  ab. 

14.  6  rs4-4  r  —  9  s  —  6.  21.  ax  4-ay  +  az  4-  bx+by-\-bz. 

22.  a#  —  a  —  ex  +  c  —  bx  +  6. 

23.  ax  —  ay  +  2  bx  —  2  5?/  +  ex  —  cy. 

24.  2  a  —  2  b  +  2  c  +  az  —  bx  +  c#  +  ay  —  by  +  cy. 

118.  III.    Type  form :  tf  +  Zab  +  V. 

a2  ±  2  aft  +  b2  =  (a  ±  6)2. 

1 .   Factor  25  x4  -  60  x2y*  +  36  y\ 

Here  a2  =  25x4,  62  =  36?/6. 

Hence  a  =  5  x2,      ft  =  6  y3, 

and  -  2  a&  =  -  2  •  5  x2  ■  6  y3  =  -  60  x2ys. 

This  product  —  60x'2y3  is  the  same  as  the  middle  term  of  the  trinomial 
which  we  are  factoring. 

Hence  that  trinomial  is  factored  by  type  form  III,  and  we  obtain  — 
25  x4  -  60  x'2y3  +  36 1/6  =  (5  x2  -  6  y3)2. 


96  ELEMENTARY   ALGEBRA 

2.    Factor  4a;2 +  2«  +  l. 
Here  a2  =  4  a;2,  62  =  1. 

We  obtain  a  =  2x,     b  =  1,  +  2  ab  =  4x. 

The  product  4  a;  is  not  the  same  as  the  middle  term  of  the  given 
trinomial. 

Hence  this  trinomial  cannot  be  factored  by  type  form  III. 

The  type  form  here  considered  enables  one   to  factor  any 
trinomial  which  is  a  perfect  square. 

EXERCISES 
119.    Factor  and  check  by  multiplying  the  factors  together: 

1.  ra2  +  2ran  +  n2.  12.    100  +  20  ft  +  ft2. 

2.  a4  -  4  a?b  +  4  b2.  13.    9  w2  —  18  w  +  9. 

3.  a4-6a262  +  9  64.  14.    1-26  6 +  169  ft2. 

4.  a2 +  24  a; +144.  15.   25  +  40a  +  16a2. 

5.  b«  +  16  +  8b\  16.   9 c2  +  54 ccPe3  +  81  d6e\ 

6.  d2  +  16d  +  64.  17.   16  n2  -  56  np  +  49 p2. 

7.  i.  +  y2-y.  18.   225  t2  - 120  ty  + 16  y2. 

8.  £a2  +  |a;  +  TV  19.   4 m2  +  64 ran  +  256 n2. 

9.  36  a2  -  12  ab3  +  &6.  20.   225  s2  -  60  st&  +  4  tw. 

10.  121e8-22e4/+/2.  21.   (a+  6)2  -  6(a  +  6)  +  9. 

11.  9y2  +  6,y  +  l.  22.    (x - yf  +  S(x - yf  + 16. 

23.  36(r  -  s)2  -  60(r  -  s)  +  25. 

24.  9(a  +  6)2-6(a  +  &)(c  +  d)+(c  +  d)2. 

25.  4-12(a-a)+9(a-a;)2. 

26.  x2  +  2xy  —  2xz  +  y2  —  2yz  +  z2. 

27.  c2  +  2  cd  +  2  ce  +  <P  +  2  de  +  e2. 

28.  u2-2a&  +  4a  +  &2-46  +  4. 

29.  4ra2  +  4ran  +  4ra  +  n2  +  2n  +  l. 

30.  g2  +  6gh  +  2g  +  9h2  +  6h  +  l. 


FACTORING  97 

120.  IV.    Type  form  :  a2  -  b2. 

a2  -  b2  =  (a  +  b)(a  -  b). 

Factor  81afy2-100z4. 

Here  a2  =  81  tfy*,  b*  =  100  z*. 

We  obtain  a  =  V8la*y2  =  9  rty,  b  =  VlOOz4  =  10  z2. 

Hence  81  z«y2  -  100  z4  =  (9  ;/:3y  +  10  z2)  (9  z*y  -  10  z2) . 

We  see  that  every  binomial  which  is  the  difference  of  two 
perfect  squares  is  composed  of  two  binomial  factors. 

Rule.     (1)  Find  the  square  root  of  each  perfect  square. 

(2)  Take  the  sum  of  the  square  roots  as  one  factor,  and 

(3)  The  first  square  root  minus  the  second  square  root  as  the 
other  factor. 

EXERCISES 

121.  Factor: 

1.  x2-y2.  7.  25c6-49d4.  13.  ft2 -169. 

2.  a2 -9.  8.  81-  121  aP.  14.  4a2-l. 

3.  1-b2.  9.  9f*-16ft  15.  16a262-25c2. 

4.  4-m4.  io.  100  p2q2 -81  u?.  16.  225  g2- 144  z*. 

5.  t^  —  25.  11.  36  a10  — 49  y*.  17.  100  u2  -  256  v*. 

6.  64  &-z2.  12.  144  v?-  25  x*.  18.  9aty*  — 121. 

19.  49c6-64*8.  29.    62-186  +  81-49s2. 

20.  289  aW  —  25  d2.  30.   d2  -  12  d  +  36  -  25 12. 

21.  {x+y)2-l.  31.    l-(a  +  6)-. 

22.  («  +  3&)2-c2.  32.    16-(x-y)2. 

23.  (a?  -  2  y)2  -  z2.  33.    9-(ra-3rc)2. 

24.  (m  +  6  ?i)2  -  4  r2.  34.    k2  —(h  +  10)2. 

25.  (7  d  -  r)2  -  25  s2.  35.    a2-62-2  6c-c2 

26.  a2  +  2a6  +  62-4c2.  =a2-(b2  +  2bc  +  c2) 

27.  x2  -  2  a*/ + 1/2  -  z2.  =  a2-(6  +  c)2. 

28.  tf-Qcd  +  ^dt-e2.  36.   t2  -  m2 +  2mn  -  n2. 


98  ELEMENTARY   ALGEBRA 

37.  49-c2-6cd-9d2. 

38.  81j*-36g*  +  12g-l. 

39.  lOOs'-ieh^-ieh-A. 

40.  (a  +  b)2-(c  +  d)2=[(a  +  b)  +  (c  +  d)][_(a  +  b)-(c  +  d)] 

=  (a  +  b  +  c  +  d)(a  +  6  -  c  —  d). 

41.  (x  +  2yf-(m-n)2.         44.  144(ra  -  a)2  —  121  (n  +  y)\ 

42.  9(c  -  df-  l(x  -  y)2.         45.  36(#  -  z)2  -  196(w  +  xf. 

43.  64(&  +  If-  49(r -  ty.       46.  c2+2cd+d2-m?+2mr-r2. 

122.  V.   Type  form  :  x2  +  gx  +  h. 

x2  +  gx  +  h=(x  +  a)(x  +  6), 
wherein  g  =  a  +  b  and  h  =ab. 

1.  Factor  a2  —  5  x  + 6. 

Here  #  =—  5,  h  =  6.     .*.  a  +  6  =—  5,  ab  =6. 

By  trial,  find  two  numbers  whose  sum  is  —  5,  and  whose  product  is  6. 
These  numbers  are  —  3  and  —  2. 

Hence  *2  -  5 x  +  6  =  (x  —  2)(x  —  3). 

2.  Factor  z2  -  2  a;  -  35. 

Here  gr  =  2,     A  =  -  35. 

That  is,  a  +  6  =  2,  a6  =  —  35. 

By  trial,  find  two  numbers  whose  sum  is  +  2  and  whose  product  is 
-35. 

These  numbers  are  +  7  and  —  5. 

Hence  x2  +  2x  -36  =(*  +  7)(*  —  6). 

3.  Factor  x2  +  5  x  —  7. 

Here  g  =  5,  ft  =  —  7. 

a  +  b  =  5,  db  —  —  1. 

We  cannot  find  two  integers  whose  sum  is  5  and  whose  product  is  —  7. 
Hence  this  trinomial  cannot  be  factored  by  this  type  form. 

ORAL  EXERCISES 

123.  Factor: 

1.  tf  +  Sx  +  S.  3.   x*  +  x-&. 

2.  a?-x-6.  4.   ^  +  9  a; +  20. 


FACTORING  99 

5.  x2-x-2Q.  21.    d2  +  7d  +  10. 

6.  x2-9z+20.  22.   d2-Ud-  15. 

7.  gi  +  x  _  20.  23.  f-  15/+  26. 

8.  a^-a;-12.  24.   ^2  +  9  gr  +  8. 

9.  ar^  9a: +  14.  25.   /t2  +  10/i-39. 

10.  a;2 -3  a; -18.  26.  h2  -  11  h  -  60. 

11.  ar' +  14  .t  + 33.  27.  A;2 -12  A; +  32. 

12.  ar*  -  16  a;  -  36.  28.  ft2  -  3  A;  -  54. 

13.  ^  +  14^+45.  29.  fc2  +  27&-90. 

14.  3^  + 14  a: +  24.  30.  P  -  25  I  + 100. 

15.  qi?  -x  —  72.  31.  m2  +  9  m  -  112. 

16.  x2  -  16  a;  +  28.  32.  rV  +  12  rs  +  20. 

17.  a2  -3a  -10.  33.  a2b2  +  16 ab  +  55. 

18.  m2  +  7m  +  12.  34.  afy2  -  an/ -  380. 

19.  a3+3a  +  2.  35.  c2^  -  12  cd  -  13. 

20.  a2 -a- 2.  36.  (a  +y)2— 7(x  +  y)-18. 

124.    VI.   Type  form :  px2  +  qx  +  r. 

px2  +  qx  +  r=  (ax  +  b)(cx  +  a*), 
wherein  p  =  ac,  q  =.  be  +  ad,  r  =  bd. 

1.   Factor  6  x2  -  x  -  12. 

Here  p  =  6,  g  =  —  1,  r  =  —  12. 

If  possible,  we  must  find  numbers  a,  6,  c,  and  d,  such  that  ac  =  6, 
6d=-  12,  be  +  ad=-  1. 

Here,  ac  =6  suggests  the  values  a  =  3,  c  =  2,  or  a  =  1,  c  =  6,  etc., 
and  bd  =  —  12  suggests  the  values  b  =  4,  d  =  —  3,  or  6  =  —  6,  d  =  2,  etc. 

Try  the  various  sets  of  values  of  a  and  c,  whose  product  is  6,  with  each 
pair  of  values  of  b  and  d,  whose  product  is  —  12,  until  a  combination  is 
found  which  satisfies  be  +  ad  =  —  1 ;  that  is,  a  combination  which  gives 
—  x  as  the  middle  term  of  the  product. 

Try  the  sets  of  values,  a  =  3,  c  =  2,  6  =  4,  d  =  —  3. 


100  ELEMENTARY   ALGEBRA 

+8x 

Examine  (3  x  +  4)  (2  x  -  3) . 

I i 

-9x 

We  see  that  the  sum  of  the  two  cross  products  +  8  x  and  —  9  x  is  —  x, 

which  is  the  middle  term  of  6  x'2  —  x  —  12. 

Hence  6  x2  -  x  -  12  =  (3x  +  4)(2  x  -  3). 

2.    Factor  8  a?2 -34  a; +  35. 

Try  a  =  8,  c  =  1,  6  =-7,  d=-5. 

-7  as 


l 1 

Examine  (8  x  —  7 )  (x  —  5) . 


-40  x 
The  sum  of  the  cross  products  —  40  x  and  —  7  x  is  —  47  x. 
But  the  middle  term  in  8  x2  —  34  x  4-  35  is  not  —  47  x. 
Hence  (8  x  —  7)(x  —  5)  are  not  factors  of  8  x'z  —  34  x  +  35. 
Try  a  =  4,  c,  =  2,  b  =  -  7,  d  =  -  5. 

-14x 


l 1 

Examine  (4x  -  7)(2x  — 5). 

I I 

-20  x 

The  sum  of  the  cross  products  —  20  x  and  —  14  x  is  —  34  x. 
—  34  x  is  the  middle  term  of  8  x2  —  34  x  +  36. 
Hence  8x2  -  34x  +  35  =(4x- 7)(2x- 5). 


EXERCISES 
125.   Factor: 

1.    Qx2+7x  +  2.  9.   10  b2-  37  b  +  30. 

2. 

3. 

4. 

5. 


6  x2  +  13  x  +  5.  10.  6  y2  -  31  y  +  35. 
12a»  +  7oj-5.  11.  6*2-  19* +  10. 

7  x2  + 11  x  +  4.  12.  3  s2  +  13  s  -  10. 
3  a;2 +  7  a +  2.  13.  24  s2  +  s  -  3. 

6.  20z2  +  33a;+7.  14.  30  y2  +  19  y  -  5. 

7.  2  a2  -  27  a  +  13.  15.  8  m2  -  22  m  -  21. 

8.  8  m2 -  31  m  +  30.  16.  3m2 -m -2. 


FACTORING  101 

17.  15m2  +  14m-8.  21.  11  a2  -  21  g  -  2. 

18.  18a2  +  9a-2.  22.  2 g*  +  gh- 15  h\ 

19.  6c2  +  35c-6.  23.  3a24-lla6-462. 

20.  6cP-13d-5.  24.  S^-Taft-Gfi2. 

126.    VII.    Type  forms  :  a3  ±  ft3. 

(a3  +  63)  =  (a  +  6) (a2  -  a&  +  &2)- 
(a3  -  63)  =  (a  -  b) (a2  +  a&  4-  &2). 

Factor  aj»  +  27. 

aa  =  a*    63  =  27  ;  a  =  Z,  6  =  3. 
.-.  x3  +  27  =  (z  +  8)(d^-8x+9). 

Again,  factor  125  m3  —  8. 

Here  a3  =  125  m3,  63  =  8 ;  a  =  5  m,  6=2. 

.-.  125  m3  -  8  =  (5  m  -  2)(25m2  +  10  n»  +  4). 

Factor : 


1. 

x3  +  y3. 

11. 

rf  +  y6. 

2. 

a?  —  y3. 

12. 

8a6-b6. 

3. 

m3  -f  1. 

13. 

1  -  27  a9. 

4. 

m3  —  1. 

14. 

m12  +  nn. 

5. 

a363  +  8. 

15. 

27  cls  +  1. 

6. 

&d3  -  27. 

16. 

8  a3n  - 1. 

7. 

64  a3  - 1. 

17. 

b9n  +  <*» 

8. 

125  W  +  216. 

18. 

tf+3/». 

9. 

8  g3  -  27  /i3. 

19. 

64  a3  -  125  ft3. 

10. 

aty»  +  343. 

20. 

216  -  27  x18. 

The  foregoing  are  the  simpler  type  forms  of  factoring  in 
algebra;  more  complicated  ones  are  given  in  more  advanced 
texts  on  algebra. 

127.  Skill  in  factoring  depends  upon  an  easy  recognition  of 
the  type  forms  and  a  facility  in  applying  correctly  the  method 
of  each  one. 


102 


ELEMENTARY   ALGEBRA 


Some  suggestions  will  be  helpful. 
I.    See  if  the  terms  contain  a  common  monomial  factor. 
If  so,  divide  the  expression  by  it,  and  keep  it  as  one  factor;  the 
quotient  will  be  the  other  factor. 

II.   Determine   to  which    type   form    the    quotient    thus 
obtained  belongs  and  factor  accordingly. 

III.   Inspect  each  factor  and,  if  possible,  resolve  it  into  new 
factors. 

TYPE  FORMS   AND  FACTORS 

128.   I.   ay+by-cy  =  y(a  +  b-c). 

II.   ax  +  ay  +  bx  +  by  =a(x+y)+b(x+y)  =  (x+y)(a+b). 

III.  a*±2ab+b*=(a±by. 

a2  -  ft2  =  (a  +  ft)(a  -  b). 

a*  +  2ab  +  b*-ci=(a+by-cl=(a+b+c)(a+b-c). 
tf  +  2  ab  +  ft2  -  c2  -  2  cd  -  d2  =  (a  +  &)2  -  (c  +  d)2 
=  (a  +  b  +  c  +  d)(a  +  b  —  c—d). 

V.   x2  +  gx  +  h  =  (x  +  a)(x  +  b) ;  g=  a  +  ft,  h  =  ab. 

VI.  px2+qx+r=(ax+b)(cx+d)  ;  p=ac,  q=bc+ad,  r=bd. 

a3  +  ft3  =  (a  +  6)(a2-  aft  +  ft8). 
a3  -  ft3  =  (a  -  ft)(a2  +  ab+  ft2). 


IV. 


VII 


REVIEW   EXAMPLES 
129.    Factor: 

1.  a?  +  2a?  +  a, 

2.  2z4  +  12z2  +  18. 

3.  9^-24^4-16^. 

4.  bz-b. 

5.  a^-16. 

6.  a2  +  6  a  -7. 


7.  a2 +  16  a +  15. 

8.  ax  —  2x—  ay  +  2y. 

9.  a2^.  —  62m  —  a2n  +  &*n. 

10.  c3  +  c2  — c  — 1. 

11.  z6  —  6^-7. 

12.  fc4-17A?  +  72. 


FACTORING  103 


13.  64  -x*.  17.  3  m4-  8  m2  -35. 

14.  h3-  7  ft2 -2Zi  +  14.  18.  y»-64. 

15.  81  ft4 -ft4.  19.  !/  +  £  +  $» 

16.  a,-2  —  2xy  +  y2-z2.  20.  x3  +  2a?  —  4  a; -8. 

21.    (a  +  6)(a2  +  62)  +  (a  +  6)(2a&). 

22.  2  a3 -14  a2 +  70 -10  a.  36.  c16-216c7. 

23.  (a  +  6)2  +  2(a  +  6)  +  l.  37.  a2  -  4  62  +  4  6c- c2. 

24.  a2  +  62  -  2  ab  -  9.  38.  a2  +  m2  -  n2  -  2  am. 

25.  c2  -  a2  +  2  a&  -  62.  39.  3  x2  -  3  ax-  +  3  x  -  3  a. 

26.  1  +  a;12.  40.  7  c2  +  20  c  +  13. 

27.  r*  +  3r  — 164.  41.  e3  +  512/3. 

28.  x2  +  llx-210.  42.  6y2-56z2  +  41yz. 

29.  a12  -  612-  43.  a2  -  12  a6  +  36  62  -  t2. 

30.  2  s2  -5s  +  2.  44.  l-9a?-30a*/-25y2. 

31.  7  a2 +  175 -70  a.  45.  9  c*  -  16  &d2  +  7  d4. 

32.  18-6x- 12x2.  46.  9  m4  -  13  m2/i2  +  4  n*. 

33.  17  w2  +  25  it,'  -  18.  47.  a?  —  b2  -  a  -  b. 

34.  9  x2*  —  25.  48.  a3  +  b3  +  a2  -  b2. 

35.  a2  -  ac  +  2  a6  -  2  6c. 

49.    4  a2  -  4  a6  +  62  -  9  c2  -  24  cd  -  16  d2. 

50.  (*  +  y)2  -  5(x  +  y)  +  6.  54.  25  z6  -61afy>  +  36  f. 

51.  a5xy  —  ab*xy.  55.  x6  +  729y6. 

52.  4  a5"  -  x\  56.  196  r%  -  z)2  -  225  s2. 

53.  49  a21  -  84  a*&»  +  36  6* 

+  4  xv  +  4  a; 

14  e  -  35. 


57.  4x2  +  3/2+  l  +  4a??/  +  4a;  +  2y. 

58.  21  c  -  5  d  +  3  cd  -  2  de  -  14  e  -  35. 

59.    (cc2  -  l)2  -  (y2  -  l)2.  60.   49y2-70yz  +  25z2. 


CHAPTER   VIII 

EQUATIONS   SOLVED   BY  FACTORING 

130.  The  degree  of  an  equation  in  one  unknown  is  the  same 
as  the  highest  power  of  the  unknown  in  the  equation. 

An  equation  of  the  first  degree  is  a  linear  equation ;  one  of 
the  second  degree  is  a  quadratic  equation ;  one  of  the  third 
degree  a  cubic  equation ;  one  of  the  fourth  degree  a  quartic 
equation ;  etc. 

ax  +  b  —  0  is  a  linear  equation. 
ax2  +  bx  -f  c  =  0  is  a  quadratic  equation. 
ax3  +  bx7-  +  ex  +  d  =  0  is  a  cubic  equation. 
ax*  -f-  bx2  +  c  =  0  is  a  quartic  equation. 

An  equation  of  second,  third,  fourth,  or  higher  degree  may 
sometimes  be  solved  by  the  methods  of  factoring  which  we 
have  learned. 

The  solution  of  such  equations  by  the  method  of  factoring 
depends  upon  the  principle  that  if  the  product  of  two  or  more 
factors  equals  zero, , at  least  one  of  the  factors  must  equal 
zero. 

131.  1.   Solve  x2  -  7  x  =  - 12. 

Transposing  —  12,  so  that  the  right-hand  member  of  the  equation  may 

be  °'  x2  -  7  x  +  12  =  0. 

Factoring,  (x  -  4)  (x  —  3)  =  0. 

Since  the  product  of  the  factors  (x  —  4)(x  —  3)  is  equal  to  zero,  one  or 
the  other  of  the  factors  must  be  equal  to  zero. 

When  x  -  4  =  0, 

we  obtain  the  root  x  =  4. 

104 


EQUATIONS  SOLVED  BY  FACTORING  105 

When  x  -  3  =  0, 

we  obtain  the  other  root,  x  =  3. 

Hence  the  given  quadratic  equation  has  the  two  roots,  3  and  4. 

Check :    If  X  =  4,  x2  —  7  x  =  —  12  becomes 

16-28  =-12. 

-  12  =  -  12. 

If  x  =  3,  x2  —  7  x  =  —  12  becomes 

9-21  =-12. 

-  12=-  12. 

2.   Solve  afJ-x  =  2x2-2. 

Transpose,  x3  -  x  -  2  x2  +  2  =  0. 

Factor,  x(x2  -  1)  -  2(x2  -  1)  =  0. 

(x2-  l)(x-2)=0. 

(x  +  l)(x-l)(x-2)  =  0. 

Ifx  +  1  =  0,  x=-l. 

If  x- 1=0,  x  =  l. 

If  x  -  2  =  0,  x  =  2.     Ans.  -  1,  1,  2. 

Check :    If  X  =  -  1,  -1  +  1=2-2,  0=0. 

If  x  =  1,  1  -  1  =  2  -  2,  0  =  0. 

If  x  =  2,  8  -  2  =  8  -  2,  6  =  6. 

Rule.  Transpose  all  the  terms  to  the  left  side  so  that  the  right 
side  may  be  zero.,  Factor  the  left  side.  Make  each  factor  con- 
taining the  unknown  equal  to  zero  and  solve  each  resulting 
equation. 

If  there  is  a  factor  that  does  not  contain  the  unknown,  divide  both 
members  of  the  equation  by  it.  Thus,  7  x2  +  21  x  —  63  =  0  has  the  factor 
7.  Dividing  both  sides  of  the  equation  by  it,  the  simplified  equation  is 
x2  +  3  x  -  9  =  0. 

» 

132.  Observe  that  a  second-degree  equation  will  produce  two 
factors,  a  £/»V(Z-degree  equation  three  factors,  and  so  on.  There 
will  be  as  many  roots  as  linear  factors  containing  the  unknown, 
and  hence  an  equation  has  as  many  roots  as  the  highest  power 
of  the  unknown  in  it. 


106  ELEMENTARY   ALGEBRA 

Some  of  the  roots  may  be  alike,  as  in 
x2  —  4  x  =  —  4. 
ar2-4a-  +  4  =  0. 
(x-2)(x-2)  =  0. 
a- 2  =  0,  a;  =  2. 
a;  -  2  =  0,  x  =  2.    ^4ws.  2,  2. 
This  is  ealled  a  double  root. 

In  solving  an  equation  care  must  be  taken  not  to  discard  a 
factor  which  contains  the  unknown  quantity ;  by  so  doing,  a 
root  will  be  lost. 

Solve   x2  —  2  x  =s  0. 

Divide  both  sides  by  x,  X  —  2  =  0 ;  x  =  2,  one  root. 
But  by  factoring,  x(x  —  2)  =  0. 

Whence  x  =  0  ;  x  =  2,  two  roots. 

EXERCISES 

133.   Solve  and  check: 

1.  a:2 -25  =  0.  13.  cx2  =  a?c. 

2.  a;2  =  36.  14.  2x2  +  6  =  7a;. 

3.  a?2  =  4a\  15.  12a^  +  a=6. 

4.  a2  =  az.  16.  25  x2  —  70  re  =  -  49. 

5.  x2  -  7  as  =  -  10.  17.  25  x2  -  25  x  +  6  =  0. 

6.  x2  —  3  a;  =  10.  18.  x2  -  ax  —  ox  +  a&  =  0. 

7.  2  x2  -  14  x  =  0.  19.  x2  —  cue  —  bx  =  0. 

8.  50^  =  15  0;.  20.  6x2— 3ax— 2cx=— ac. 

9.  x2  +  x  =  72.  21.  x3  =  4x. 

10.  x2  — 2ax  =  — a2.  22.   x3  +  8  =  2x2  +  4 x. 

11.  2  a2  =  3  a.  23.    x3-3x2  +  2x  =  0. 

12.  foe2  —  ax  =  0.  24.   x3  +  20x  =  9x2. 


EQUATIONS  SOLVED  BY  FACTORING  107 

PROBLEMS 

134.  Solve  the  following  problems  and  see  which  roots  of 
the  equations  satisfy  the  conditions  of  the  problem : 

1.  What  number  is  it  whose  square  plus  the  number  itself 
equals  42  ? 

2.  The  product  of  a  certain  number  decreased  by  2  and 
the  same  number  decreased  by  3  equals  20.     Find  the  number. 

3.  If  to  the  square  of  a  certain  number,  the  number  itself 
and  5  be  added,  the  result  will  be  115.     Find  the  number. 

4.  If  a  certain  number  be  subtracted  from  19  and  21,  in 
turn,  the  product  of  the  remainders  will  be  80.  What  is  the 
number? 

5.  The  difference  between  two  numbers  is  3  and  the  sum 
of  their  squares  is  89.     What  are  the  numbers  ? 

6.  A  rectangular  field  is  5  rd.  longer  than  it  is  wide.  If 
each  dimension  be  increased  by  1  rd.,  its  area  would  be  50  sq. 
rd.     What  is  its  present  area  ? 

Let  x  rd.  =  the  width. 

Then  (x  +  5)  rd.  =  the  length. 

(x  -f  1)  rd.  =  the  width  increased  by  1  rd. 
(x  +  6)  rd.  =  the  length  increased  by  1  rd. 
We  obtain  the  equation  (x  -f-  l)(x  -f  6)=  50. 
x2  +  7  x  +  6  =  50. 
x2  +  7  x-U=0. 
(x-4)(x  +  ll)  =  0. 
The  roots  of  the  equation  are  +  4  and  —  11. 

We  discard  the  root  —  11  as  being  inapplicable  to  our  problem,  on  the 
ground  that  we  do  not  speak  of  a  rectangle  having  a  negative  length  and 
a  negative  breadth  ;  the  problem  involves  the  condition  that  only  the 
positive  roots  shall  be  used. 

7.  If  the  length  of  a  certain  square  be  increased  3  ft.  and 
the  width  decreased  2  ft.,  the  area  would  not  be  changed. 
What  are  the  dimensions  of  the  square  ? 


108  ELEMENTARY   ALGEBRA 

8.  By  how  many  inches  must  the  length,  38  in.,  and  the 
width,  22  in.,  of  a  rectangle  be  increased,  in  order  to  increase 
the  area  by  189  sq.  in.  ? 

9.  How  may  $  84  be  divided  among  a  number  of  persons 
so  that  each  person  shall  receive  $  5  more  than  the  number  of 
persons  ? 

10.  Is  the  difference  of  the  squares  of  two  consecutive  num- 
bers even  or  odd  ?     Prove  your  answer. 

11.  The  sum  of  the  squares  of  two  consecutive  numbers  is 
85.     What  are  the  numbers  ? 

12.  The  sum  of  the  squares  of  two  consecutive  odd  numbers 
is  202.     What  are  the  numbers  ? 

13.  A  schoolroom  contains  48  seats,  and  the  number  in  a 
row  is  4  less  than  twice  the  number  of  rows.  How  many  seats 
in  a  row  ? 

14.  If  a  cubical  box  has  its  dimensions  increased  by  1,  2,  3 
inches  respectively,  the  capacity  of  the  box  has  been  increased 
93  cu.  in.     What  are  the  dimensions  ? 

15.  A  rectangular  yard  has  a  fence  around  it  measuring 
100  ft. ;  the  area  of  the  yard  is  600  sq.  ft.  How  does  its 
width  compare  with  its  length  ? 

16.  The  surface  of  a  cube  contains  384  sq.  in.  Find  its  di- 
mensions. 

The  altitude  of  a  triangle  is  its  height,  or  the  length  of  a  perpendicular 

line  from  its  vertex  to  its  base. 

The  area  of  a  triangle  equals  one 
half  its  altitude  multiplied  by  its  base. 
Area  =  \ab. 

17.    If  the  area  of  a  triangle  is 
FlG  19  36  sq.  in.  and  the  base  is  12  in., 

find  the  altitude. 
18.    The  base  of  a  triangle  is  3  times  the  altitude  and  the 
area  is  24  sq.  ft.     Find  the  base. 


EQUATIONS  SOLVED  BY  FACTORING 


109 


If  the  perimeter  is  72  in., 


19.  The  area  of  a  triangle  is  35  sq.  ft.  and  the  base  is  3  ft. 
more  than  the  altitude.     Find  the  altitude  and  the  base. 

The  hypotenuse  of  a  right  triangle  is  its  longest  side.     The  square  on 
the  hypotenuse  is  equal  to  the  sum  of 
the  squares  on  the  legs. 

20.  One  leg  of  a  right  triangle 
is  3  ft.  longer  than  the  other,  and 
the  area  is  54  sq.  ft.  Find  each 
leg  and  the  hypotenuse. 

21.  The  sides  of  a  certain  right 
triangle  are  in  the  ratio  3:4:5. 
what  is  its  area  ? 

22.  The  diagonal  of  a  square  is  50  in.     What  is  its  perim- 
eter ? 

23.  The  diagonal  of  a  rectangle  is  640  in.     If  the  length  of 
the  rectangle  is  three  times  its  width,  what  is  its  area  ? 

24.  The  area  of  a  rectangle  is  48  sq.  ft.     If  its  length  is 
2  ft.  more  than  the  width,  what  will  its  diagonal  be  ? 

The  area  of  a  trape- 
zoid is  equal  to  one 
half  the  altitude  times 
the  sum  of  the  bases. 

Area=«(»+y>. 

2 

25.    The     alti- 
tude of  a  trapezoid 
is  12  in.,  one  base 
is  3  in.  shorter  than 
How  long  is  the  other 


b' 
Fig.  21 

the  other;    the  area  is  162  sq.  in. 
base? 


26.  One  base  of  a  trapezoid  is  2  ft.  longer  than  the  other 
and  the  altitude  is  1  ft.  longer  than  the  longer  base ;  the  area 
is  120  sq.  ft.     Find  the  bases  and  the  altitude. 


110  ELEMENTARY   ALGEBRA 

The  length  of  a  circle  (circumference),  C  =  2  irR  (wherein  tr(pi)  is  a 
Greek  letter  standing  for  3.1416,  or  %?-,  approximately). 
The  area  included  by  a  circle  A  =  irR2. 

27.  If  a  circle  is  88  in.  long,  what  is  its  radius  ? 

28.  Over  how  much  ground  may  a  horse  graze  if  tied  to  a 
stake  by  a  30-foot  rope  ? 


FRANCOIS   VIETA 


Vieta's  Notation  Modern  Notation 

1  C-8  #  +  16iVaequ.  40;  Xs  -  8  a;2  +  16  a;  =  40. 

C  stands  for  cubus  or  "  cube  "  of  the  unknown. 
Q  stands  for  quadratus  or  "square  "  of  the  unknown. 
N  stands  for  numerus  or  the  unknown  itself. 
aequ.  stands  for  aeqnalis  or  equal. 


CHAPTER   IX 

MULTIPLICATION   AND   DIVISION   OF   POLYNOMIALS 
REVIEW  OF   MULTIPLICATION 

135.  Arrange  both  polynomials  that  are  to  be  multiplied  to- 
gether, according  to  the  descending  or  ascending  power  of 
some  letter.  Multiply  each  term  of  one  polynomial  by  each 
term  of  the  other,  and  add  the  partial  products. 

Thus,  find  the  product  of : 

1.  2  a2  +  3  a  -  5  and  3  a?  —  2  a  +  1. 

2.  3  a?  -  a2  +  4  a  -  2  and  a2  +  a  -  3. 

3.  3  x2  —  x  -f-  4  and  2  x2  +  3  x  —  5. 

4.  x2  -f-  a;*/  -+-  y2  and  x  —  y. 

5.  2  a;2  -  5  a:?/  +  2  y2  and  2  a  —  3  y. 

6.  a?4  —  x2  +  1  and  x4  4-  x2  +  1. 

7.  1  -  a  -  2  a2  +  a4  and  1  —  2  a  +  a2. 

8.  c6  -  2  c3  +  1  and  c2  -  2  c  +  1. 

9.  x  —  y  and  afy  +  a;^3  +  y*  +  x2y2  +  x*. 

10.  1  +  5  a?  -  6  a4  and  3  a2  +  1  —  a. 

Simplify : 

11.  (ra4-9m2  +  12ra-4)(ra2  +  3m-2). 

12.  (3r3-2r  +  7)(5r4-f3r2-l). 

13.  (a?  +  a2b  +  ab2  +  b3)(a  -  b)(a*  +  b*). 

14.  (x2  +  y2  -\-z2  —  xy  —  xz  —  yz)(x  +  y  -\-  z). 

15.  (x2  -|-  3  x  -  2)(4  x2-oa;  +  1)(2  x2  -  x  -  3). 

ill 


112  ELEMENTARY   ALGEBRA 


DIVISION  OF  POLYNOMIALS 


136.  We  learned  (§  76)  to  divide  a  polynomial  by  a  bino- 
mial. This  process  may  be  extended  to  cases  in  which  the 
divisor  is  itself  a  polynomial. 

(1)  Divide  §xi-5x3-lx2  +  %x-2\)y2x2-3x  +  l. 


6 a4- 5 as1—    7^  +  8^-2 
6^-9^+    3.r2 


2  x2  -  3  x  +  1 


3  x*  +  2  x  -  2 


Ax3-  10  a;2  +  8  a;  -2 
4a3-    6  a;2 +  2  a; 

-  4a;2  +  6a;- 2 

-  4  a;2 +  6  a; -2 

There  is  no  remainder ;  the  division  is  exact.     The  quotient 
is3z2  +  2z-2. 

(2)  Divide  16  +  5  a4  -  3a3  -  32  a  +  36  a2  by  a2  +  4  -  4  a. 
16 -32  a +  36  a2-    3a3  +  5a4|4  -  4a  +  a2 

16-16a  +    4  a2 |4-4a  +  4a2 

-16  a +  32  a2-    3  a3 +  5  a4 
-16 a +  16  a2-    4a3 

16  a2  +      a3  +  5  a4 
16a2 -16  a3 +  4  a4 

17  a3  +  a4 

There  is  a  remainder,  and  the  division  is  "not  exact."     The 

quotient  is  17a3+a4 

4  —  4a  +  4a2  + 


4  —  4a  +  a2 


137.  Emphasis  must  be  placed  upon  the  necessity  of  arrang- 
ing both  polynomials,  and  each  remainder,  according  to  the  as- 
cending or  the  descending  power  of  some  letter. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of  the 
divisor  for  the  first  term  of  the  quotient. 

Multiply  the  entire  divisor  by  the  first  term  of  the  quotient. 

Subtract  the  product  from  the  dividend. 

Treat  the  remainder  as  a  new  dividend  and  proceed  as  before. 


POLYNOMIALS  113 

To  check  an  example  in  division : 

I.  Multiply  the, quotient  by  the  divisor,  add  the  remainder 
to  this  product.     The  result  should  be  the  dividend ;  or 

II.  Substitute  particular  numbers  for  the  letters  and  divide 
the  resulting  values. 

To  illustrate  the  second  method  of  checking :  in  our  first  division,  let 
x  =  2  ;  then  6  x*  -  5  x8  -  7  x2  +  8  x  -  2  =  42. 

2x2-3x+l  =  3. 
3x2  +  2x-2  =  14. 
Then,  42  +  3  =  14. 

Hence  the  division  is  correct.  In  this  mode  of  checking,  care  must  be 
taken  to  select  numbers  for  the  letters  which  will  not  reduce  the  divisor 
to  zero. 

EXERCISES 
138.    Divide  and  check  : 

1.  x4  -  9  x3  +  19  x2  -  25  x  +  6  by  x2  -  2  x  +  3. 

2.  6  a4-  a3b  -  10  a2b2  +  31  ab3  -  20  b*  by  3  a2  +  4  ab  -  5  b2. 

3.  8  a4  +  8  a3b  -  28a262  -fc  53 a&3-  21  &4  by  4a2  -  Gab  +  lb2. 

4.  x5-x4  +  4x3  —  3x2  +  5x  —  ebyaf  +  x—  2. 

5.  a3  -  3  a2b  +  3  ab2  -  b3  by  a2  -  2  ab  +  b2. 

6.  10a5+lla4-39a3  +  42a2-17a+5by  2a3+3a2-7a+5. 

7.  r8  +  r1  —  is  —  r5  +  r3  -f  r2  —  1  by  r4  —  r2  +  1. 

8.  a4  +  a262  +  &4  by  a2  +  b2  -  ab. 

9.  x5  —  yh  by  x  —  y. 

10.  a3  +  b3+  c3  -  3  abc  by  a  +  6  +  c. 

Simplify : 

11.  (x2  +  2  x  +  2)(x2  -  4  x  -  5)  -=-  (x  -  5). 

12.  (x  +  y)2(x2  —  xy  +  y2). 

13.  [(a3  +  63)  +  (a2  -  ab  +  ft2)]  [a2  +  a6  +  62]. 

14.  (a  -  6)3(a  +  6)2  --  (a2  -  b2). 

15.  [(x  +  2/)3-(x-t/)3]-(3x2  +  i/2). 


CHAPTER  X 

SQUARE  ROOT 

139.  The  square  root  of  an  expression  is  one  of  the  two 
equal  factors  of  the  expression. 

Since  (o  +  ft)2  =  a2  +  2  aft  +  ft2, 

Va2  +  2  ab  +  ft2  =  a  +  b. 

Thus  it  is  a  simple  matter,  by  inspection,  to  extract  the 
square  root  of  a  trinomial,  which  is  seen  to  be  of  the  form 
a2+2aft  +  ft2. 

Or  we  may  find  the  square  root  in  this  way : 

a2+2aft  +  &2    \a  +  b 
a2 


Trial  divisor,  2  a  2  ab  +  ft2 

Complete  divisor,     2  a  +  ft  2  aft  +  &2 

Thus  the  square  root  is  a  +  6. 

The  study  of  this  simple  case  enables  one  to  devise  a  rule 
which  is  applicable  to  more  complicated  cases.  The  procedure 
is  as  follows : 

I.  Extract  the  square  root  of  the  first  term  and  subtract  its 
square  from  the  polynomial,  leaving  2ab  +  ft2. 

II.  "We  see  in  2  aft  the  factor  ft  which  we  know  is  the  second 
term  of  the  root.  In  cases  when  6  is  not  known,  we  see  that  ft 
may  be  obtained  by  dividing  2  aft  by  2  a.  We  call  2  a  the  trial 
divisor.  Since  a  is  the  part  of  the  root  already  found,  we  see 
that  the  trial  divisor  is  double  the  root  already  found.  After  ft 
has  been  found,  add  it  to  the  trial  divisor  and  we  have  2  a  +  6, 
the  complete  divisor. 

114 


SQUARE  ROOT  115 

III.    Multiply  2  a  +  b  by  &,  and  subtract. 
This  process  may  now  be  extended  to  finding  the  square  root 
of  any  polynomial. 

140.    The  extraction  of  the  square  root  of  49  cc4— 42ar,+79x2 
—  30  x  +  25  is  as  follows  : 

49x4-42x8+79x2-30x-|-25    |7x2-3x+5 
49  x4 


-42x8  +  79x2— 30x  +  25,  1st  remainder 
-42x8+  9x2 


+70x2-30x+25,  2d  remainder 
+70x2-30x+25 


1st  trial  divisor,  2(7  x2)  =  14x2 

1st  complete  divisor,    14  x2— 3  x 

2d  trial  divisor,  2(7  x2— 3x)  =14x2—  6x 

2d  complete  divisor,  14  x2— 6x+5 

0  3d  remainder 

The  process  is  as  follows  : 

(1)  The  square  root  of  49  x4  is  7  x2.  Here  a  =  7  x2 ;  2  a  =  14  x2,  the 
first  trial  divisor. 

(2)  The  first  term  in  the  first  remainder  is  —42  x3.     Take  —42  x3=2  ab. 

(3)  To  find  b,  divide  -  42  Xs  by  14  x2.     We  obtain  b  =  -  3x. 

(4)  Multiply  the  first  complete  divisor  by  —  3  x  and  subtract.  The 
second  remainder  is  what  is  left  after  subtracting  (7x2  —  3x)2  from  the 
given  polynomial. 

(5)  Proceed  with  the  second  remainder  as  you  did  with  the  first.  Let 
7  x2  —  3  x  be  the  new  value  of  a. 

(6)  Divide  the  first  term  of  the  second  remainder  by  the  first  term  of 
the  second  trial  divisor  ;  you  obtain  +  5. 

(7)  Multiply  the  second  complete  divisor  by  +  5,  and  subtract.  The 
third  remainder  is  what  is  left  over  after  subtracting  (7x2  — 3x  +  5)2 
from  the  given  polynomial.  As  this  remainder  is  zero,  (7  x2  —  3x  +  6)2 
is  equal  to  that  polynomial.  • 

Hence  7  x2  —  3  x  +  5  is  the  required  square  root. 

EXERCISES 
141.    Find  the  square  root  of : 

1.  a4-4a3  +  6a2-4a  +  l. 

2.  x4-6x3  +  13x2-12a  +  4. 

3.  c5  +  c4-26c3  +  6c54-8c  +  10c2  +  l. 

4.  4x8-12a6-15z4  +  36x2  +  36. 


116  ELEMENTARY   ALGEBRA 

5.  25  a*  -  20afy  +  34afy2  -  32a*y  +  17  aty*  -  12 xy*  +  4?/6. 

6.  9  c4^2  -  18  c?d3  -  3  c2d4  +  12  cdb  +  4  d6. 

7.  m4n4  +  16  r*  +  mWr2  —  6  ra3tt3r  +  24  mnr3. 

.      2ar»  ,  x2 

n     a2      ab  ,       ,  62      2  6  ,   i 
9.    r--  +  a  +  --y+l. 

10.   4x2-20o;--  +  t^  +  27. 
a:      4a;2 


SQUARE  ROOT  OF  ARITHMETICAL  NUMBERS 

142.  I2  =  1,  21  =  4,  ...  92  =  81,  102  =  100,  902  =  8100,  1002 
=  10,000,  10002  =  1,000,000. 

This  is  sufficient  to  show  that  a  perfect  square  consisting  of 
two  digits  has  one  digit  in  the  root ;  one  consisting  of  three  or 
four  digits  has  two  digits  in  the  root;  one  consisting  of  Jive  or 
six  digits  has  three  digits  in  the  root ;  and  so  on.  In  other 
words,  the  square  of  a  number  has  twice  as  many  digits,  or 
one  less  than  twice  as  many,  as  the  number  itself. 

Hence,  to  find  the  square  root  we  must  separate  the  number 
into  periods  of  two  digits  each,  towards  the  left  and  right  from 
the  decimal  point.  The  period  farthest  to  the  left  may  have 
one  or  two  digits,  whereas  the  one  farthest  to  the  right  must 
have  two,  a  cipher  being  annexed,  if  necessary,  to  complete  it. 

The  square  root  has  therefore  one  digit  corresponding  to  every 
period  in  its  square.  Separating  the  square  number  into 
periods  enables  one  to  find  one  digit  in  the  root  at  a  time. 

This  is  seen  more  clearly  if  we  consider  that  the  square  of 
any  number  of  tens,  say  4  tens  (40),  ends  in  two  ciphers 
(1600)  ;  hence  the  two  digits  on  the  right  are  not  needed  to 
find  the  tens'  digit  (4),  and  are  set  aside  until  the  units'  digit 
is  to  be  found. 

Likewise,  the  square  of  any  number  of  hundreds,  say  4 


SQUARE  ROOT  117 

hundreds  (400),  ends  in  four  ciphers  (160,000),  and  all  of  these 
may  be  set  aside  until  the  hundreds'  digit  is  found  and  they 
are  needed  for  the  tens'  and  units'  digits. 

After  pointing  off  the  number  into  periods,  the  method  is 
similar  to  the  one  used  for  polynomials.  The  procedure  is 
based  on  the  formula  a2  +  2  ab  +  62  =  (a  -4-  b)2. 

When  dividing  by  the  trial  divisor,  exclude,  for  brevity,  the  right-hand 
digit  in  the  remainder. 

The  reason  for  this  is  seen  in  the  example  which  follows  ;  the  trial 
divisor  2  is  equal  to  20  of  the  next  lower  units.  Now  12-^-2  gives  the 
same  digit  in  the  root  as  127  -*■  20. 

Find  the  square  root  of  2272254.76. 

2272254.76 | 1507.4 

1 
1st  trial  divisor  =  2(1)  =  2      1127  1st  remainder 

1st  complete  divisor  =  25         [125 
2d    trial  divisor  =  2(15)  =  30  _ 


3d    trial  divisor  =  2(150)  =  300 
3d    complete  divisor  =  3007 
4th  trial  divisor  =  2(1507)'=  3014 
4th  complete  divisor  =  30144 


22254  2d  remainder 

21049 


120576        3d  remainder 
120576 

Ans.  1507.4. 
When  the  2d  trial  divisor  (30)  will  not  go  into  22,  a  cipher  is  placed 
in  the  root ;  the  next  trial  divisor  becomes  2(150)  or  300  and  the  next 
period  is  brought  down.     Then  2225  is  divided  by  300. 

143.  The  square  root  of  a  fraction  may  be  found  by  taking 
the  square  root  of  the  numerator  and  of  the  denominator 
separately,  or  by  reducing  the  fraction  to  an  equivalent  decimal 
and  taking  the  square  root  of  the  decimal.  Unless  the 
denominator  of  the  fraction  is  a  perfect  square,  the  second 
method  of  procedure  is  far  better. 

EXERCISES 

144.  Find  the  square  root  of : 

1.  3249.  3.   51529.  5.   248004. 

2.  6084.  4.   93636.  6.   118.1569. 


118  ELEMENTARY   ALGEBRA 

7.  1010025.  11.   .01522756. 

8.  10.4976.  12.   1459.24. 

9.  .00564001.  13.   420.496036. 
10.  64.1601.                                   14.   7753607.3209. 

Find,  to  three  decimal  places,  the  square  root  of : 

15.  2.  16.    3.  17.    5.  18.    f.  19.    f.  20.    |. 


CHAPTER  XI 

GRAPHS  OF  SIMULTANEOUS  EQUATIONS 

145.  The  plotting  of  linear  equations  of  the  form  ax  +  by 
=  c  was  explained  in  §  90. 

In  the  equation  x  +  Oy  =  2,  y  may  have  any  value ;  but  x 
must  equal  2.  Consequently  such  points  as  (2,  0),  (2,  —  3), 
(2,  2),  (2,  5),  (2,  -  1),  (2,  -  2)  would  lie  on  the  graph.  Thus 
the  graph  must  be  a  line  parallel  to  the  y-axis,  two  units  to  the 
right. 

EXERCISES 

146.  Represent  equations  1-8  graphically : 

1.  a  =  4.  3.   x=—7.        5.   x  =  0.  7.   y  =  ±2. 

2.  y  =  —  1.      4.   y  =  3.  6.   ?/  =  0.  8.   x=±3. 

9.    Is  the  point  (3,  4)  on  the  graph  of  the  equation  2  x  —  y 
=  7?     Can  you  tell  without  making  a  graph  of  the  equation  ? 

10.  Can  you  tell  without  making  a  graph  of  an  equation 
whether  the  graph  goes  through  the  origin  or  not  ? 

11.  Which  of  the  equations  1-8  above  go  through  the 
origin  ? 

12.  What  would  you  note  about  the  graphs  of  the  equations 
2x-3y  =  6aind2x-3y  =  12? 

13.  Do  x  +  y  =  7  and  2x  +  2y  =  14:  stand  for  two  distinct 
graphs  or  only  one  ?     Why  ? 

14.  Can  you  tell  by  looking  at  a  linear  equation  whether  its 
graph  is  parallel  to  the  x-axis,  or  to  the  y-axis,  or  parallel  to 
neither  of  them  ?     How  ? 

15.  Why  are  the  equations  in  Ex.  12  called  "  inconsistent "  ? 
Will  the  same  finite  values  of  x  and  y  satisfy  both  of  them  ? 

119 


120 


ELEMENTARY   ALGEBRA 


16.    Will  the  same  values  of  x  and  y  satisfy  both  equations 
in  Ex.  13  ? 


TWO  EQUATIONS  IN  X  AND    Y 

147.  The  solution  of  a  pair  of  linear  equations  may  be 
obtained  by  drawing  their  graphs,  referred  to  the  same  pair  of 
axes. 

For  instance:  x  +  2y  =  —  4,  (1) 

3  x  -  y  =  9.  (2) 

In  (1),  if  x  =  0,   y  =  —2. 
x  =  —  4,   y  =  0. 


Y 

1 

I 

r 

jj 

~t_ 

5j 

// 

"  v?r 

j| 

>^                                                                                                                                      j_ 

■*»>L                                                                    r 

^e^s,                                                          -I 

tf^*-^.                                                    1 

"      3    ^c4  ^3    -2     -I       C              "     f~                                            X" 

vs                           _J       2^jB      <       It. 

*"*!V                                          ' 

^«._                          jl 

^w        7 

..«    s;5fe:31 

3            ^/^V 

.            T        ^^ 

r              ^"^^ 

"7                       "-v. 

T                             ^^ 

f~                                    ^^ 

1/                                             ^^ 

T                                                        ^ 

r 

"J 

/ 

r~ 

1 

z 

Fig.  22 


GRAPHS   OF  SIMULTANEOUS  EQUATIONS       121 

Plot  these  points  and  draw  the  graph. 

In  (2),  if  a;=0,   y  =  -9. 
a;  =  3,   y  =  0. 

Plot  these  points  and  draw  the  graph. 

The  graphs  intersect  at  point  P,  whose  coordinates  are  2,  —  3.  Since 
two  straight  lines  can  intersect  in  only  one  point,  there  can  be  only  one 
pair  of  values  of  x  and  y  which  will  satisfy  two  linear  equations. 

Test  these  coordinates  by  substituting  them  in  both  equations. 

Let  x  =  2,   y  =  —  3  ;  x  +2y  =  —  4   becomes  —  4  =  —  4 ;  3x  —  y  =  9 
becomes  9  =  9. 

Hence  x  =  2,  y  =  —  3  satisfy  both  equations. 

What  would  you  infer  if  the  graphs  were  parallel  ? 
What  would  you  infer  if  the  graphs  were  identical  ? 

EXERCISES 

148.  Plot  the  following  pairs  of  equations,  each  pair  on  the 
same  set  of  axes ;  determine  the  coordinates  of  the  intersection, 
and  test  the  result  in  each  pair  of  equations : 

1.  x  +  y  =  3,  7.    x  +  y  =  7, 
x  +  4y  =  9.  %  =  y- 

2.  x  —  y  =  4,  8.    x  —  2y  =  l, 
2x+y  =  U.  x  =  2\y. 

3.  x  —  y  =  5,  9.    x  —  y  =  1, 

x  +  y  =  9.  y  =  —  2. 

4.  x  +  2y  +  3  =  0,  10.   3x-2y  =  0, 
2x-3y-S  =  0.  2x  +  3y  =  0. 

5.  3a; -4?/ =  —  18,  11.   as  =  -4, 
4a  +  37/  =  l.  y  =  3. 

6.  2x-y  =  3,  12.    4x- 5^  =  20, 
4aj—  2y  =  6.  4x  —  5y  =  40. 

13.  What  is  the  equation  of  the  a-axis  ? 

14.  What  is  the  equation  of  the  y-axis  ? 


122 


ELEMENTARY   ALGEBRA 


PRACTICAL  APPLICATIONS  OF  GRAPHS* 

149.    Draw  a  graph  for  changing  prices  expressed  in  dollars- 
per-yard  to  prices  expressed  in  francs-per-meter. 

1  yard  =  .915  meter,  1  dollar  =  5.18  francs. 
Hence  $  1  per  yard  =  5.18  fr.  per  .915  meter. 

If  .915  meter  cost  5.18  fr.,  then  1  meter  costs  5.18  fr.  -=-  .915  =  5.66+fr. 
Hence  1  dollar-per-yard  =  5.66  fr.  per  meter,  nearly. 
Let  x  =  francs-per-meter, 

and  y  =  dollars-per-yard. 

Then  x  =  5.66  y. 

When  y  =  0,  then  x  —  0  ;  these  values  determine  the  point  O  in  Fig.  23. 
When  y  =  5,  then  x  =  28.3  ;  these  values  determine  the  point  A. 


I6 

&4 

|   3 
§    2 


|::::::::::::::::!:::::::|:::::::: 

X :::"::::::::::::::::::::  i ::::::: : 

:::::::|1L[.., 

B 

-\    *&*- 

I + 

oij-^i^ 

||11111I1II11I1I11|I|1P%^ 

*~        --+H- 

f           T    "--jSSegELi _ 

1 

z^ss&c^- 

■  :■      i 

■^Zr^A                         :..:::: 

~z   z    'D 

x 

O  5  10  15  20  25 

Francs  -  per  -  meter 

Fig.  23 

The  line  OA  is  the  required  graph.  To  reduce  4  dollars-per-yard  to 
francs-per-meter,  pass  from  B  to  C,  and  from  C  to  D.  The  answer  is 
22.6  francs-per-meter,  nearly. 


EXERCISES 
150.    By  inspection,  change  the  following  to  francs-per-meter : 

1.  $2  per  yard.         3.    $  6  per  yard.         5.    $4|  per  yard. 

2.  $3  per  yard.         4.   $5^  per  yard.       6.    $3^  per  yard. 

*  May  be  postponed  for  the  present. 


GRAPHS  OF  SIMULTANEOUS   EQUATIONS       123 


Change  the  following  to  dollars-per-yard : 

7.  5  fr.  per  meter.  10.    20  fr.  per  meter. 

8.  10  fr.  per  meter.  11.    25  fr.  per  meter. 

9.  15  fr.  per  meter.  12.    27  fr.  per  meter. 

13.  Given  1  mile  =  1.61  Kilometers,  and  1  dollar  =  5.18 
francs,  construct  a  graph  for  changing  dollars-per-mile  to  francs- 
per-Kilometer. 

14.  Draw  a  graph  for  changing  miles-per-hour  to  feet-per- 
minute. 

Hint.  1  mile-per-hour  —  5280  feet  per  60  minutes  =  5280  -~-  60  feet- 
per-minute. 

15.  Draw  a  graph  for  changing  rods-per-minute  to  feet-per- 
second. 

16.  A  freight  train  starts  from  Denver  at  noon,  running  at 
the  rate  of  20  miles  an  hour.  Four  hours  later  a  passenger 
train  starts  in  the  same  direction  at  the  rate  of  40  miles  per 
hour.  Draw  a 
graph  showing  the 
distances  of  the 
trains  from  Den- 
ver during  the 
first  ten  hours. 

Let  x  =  the  num- 
ber of  hours  after 
12  M,  and  y  =  num- 
ber of  miles  traveled. 

(a)  For  the  freight 
train,  y  =  20  x. 

When  x  =  0,  then 
y  =  0  ;  these  values 
determine  the  point 
O  in  Fig.  24. 

When  x  =  10,  then  01234567 

y  =  200  ;  these  values  Hours  after  12  M, 

determine  the  point  A.  Fio.  24 


y 

200 

A 

180 

160 

9 

>   140 

c 

Q  120 

e 
2  ioo 

/ 

2  80 

60 

40 

20 

u/ 

X 

8      9      10 


124 


ELEMENTARY   ALGEBRA 


The  line  OA  shows  the  distances  traveled  in  different  times. 
(6)  For  the  passenger  train,  y  =  40(a;  —  4). 
When  x  =  4,  then  y  =  0  ;  these  values  determine  the  point  B. 
When  x  =  7,  then  y  =  120 ;  these  values  determine  the  point  C 
The  line  BG  shows  distances  traveled  by  the  passenger  train. 

(c)  How  far  from  Denver  is  each  train  after  4 J  hr.  ?  5  hr.  ? 
8J  hr.  ? 

(d)  How  far  from  Denver  does  the  passenger  train  overtake  the  freight 
train  ? 

(c)  At  what  time  does  it  overtake  the  freight  train  ? 

17.  A  starts  off  on  a  bicycle  at  7  miles  an  hour.  Two  hours 
later  B  rides  in  the  same  direction  at  10  miles  an  hour.  Draw 
a  graph  and  tell  from  it  how  far  apart  the  two  men  are  after 
3  hr.,  5  hr.,  6^  hr.  Where  and  when  does  the  second  man 
overtake  the  first  ? 


18.  A  train  leaves  Denver  for  Colorado  Springs  at  8  a.m., 
traveling  30  miles  an  hour.  At  8  :  30  a.m.  a  second  train  leaves 
Colorado  Springs  for  Denver,  traveling  40  miles  an  hour.  The 
two  stations  are  80  miles  apart.  Draw  a  graph  and  tell  from  it 
how  far  apart  the  trains  are  at  9  o'clock,  9  :  30  o'clock,  10  o'clock. 

Where  and  when  do  the 
trains  meet  ? 

Let  x  =  number  of  hours 
after  8  a.m.,  and  y  =  num- 
ber of  miles  from  Denver. 

(a)  The  point  O  marks 
the  time  and  place  where 
the  first  train  starts.  Ex- 
plain. 

(6)  The  point  A  marks 
the  time  and  place  where  the 
second  train  starts.  Ex- 
plain. 

(c)  Explain  how  the  line 

i  ^2«am      3  05  is  drawn. 

Hours  after  8  A.M.  ,_   __.,..  .         ..     ,. 

(d)  Explain  how  the  line 

Fig.  25  AG  is  drawn. 


V 

80 

K 

B 

w 

70 

£ 

\ 

y 

60 

t\ 

A 

/ 

%; 

yy 

50 

V 

r*N> 

o 

\< 

y 

40 

•*  ; 

/ ( 

Sj 

30 

% 

V 

20 

^N 

tp 

1 

10 

% 

X 

GRAPHS   OF  SIMULTANEOUS  EQUATIONS       125 


19.  A  man  cycles  from  A  to  B  at  8  miles  an  hour,  and  re- 
turns at  10  miles  an  hour.  If  he  takes  4  hr.  to  go  there  and 
back,  find  the  distance  from  A  to  B.  Draw  a  graph  showing  how 
far  he  was  from  B  after  1  hr.,  2  hr.,  3  hr.    When  was  he  at  B  ? 

20.  A  boy  begins  work  with  a  weekly  wage  of  $9  and 
receives  an  increase  of  25^  every  week.  Another  boy  starts 
with  a  weekly  wage  of  only  $6,  but  receives  an  increase  of  50^ 
every  week.  Draw  a  graph  which  shows  the  wage  of  each  at 
the  beginning  of  every  week,  for  20  weeks.  When  will  their 
wages  be  the  same  ? 

Let  x  weeks  =  the  time, 

and  y  dollars  =  the  wage. 

(a)  After  x  weeks  the  wage  of  the  first  boy  is  y  =  9  +-. 

4 

When  £=0,  y—9 ; 
these  fix  the  point  A. 

When  x  =  20, 
y  =  14  ;  these  fix  the 
point  B. 

The  line  AB  shows 
the  wage  of  the  first 
boy  for  every  week. 

(6)  After  x  weeks 
the  wage  of  the  sec- 
ond boy  is  y  =  6  +  -  • 

When  x=0,  y=Q  ; 
these  fix  point  C. 

When  x  =  20, 
y  =  16  ;    these  fix  D. 

The  line  CD  shows 
the  wage  of  the  second 
boy  for  every  week. 

(c)  During  which  week  are  the  wages  the  same  ? 

(d)  State  the  difference  in  wage  after  7,  10,  13,  and  19  weeks. 

21.  Fred  has  $7  in  the  bank  and  adds  to  this  75^  a  day. 
David  has  $5  in  the  bank  and  adds  to  this  $1.25  a  day. 
Draw  a  graph  showing  the  amounts  for  the  first  7  days. 


v 

[) 

15 

10 
A 

C 

X 

10 

Weeks 
Fig.  26 


CHAPTER   XII 

SIMULTANEOUS   LINEAR  EQUATIONS 

151.  In  a  linear  equation  in  one  unknown,  that  unknown 
can  have  just  one  value.     Thus  2  x  =  8,  x  =  4. 

In  a  linear  equation  in  two  unknowns,  one  unknown  may 
vary  at  pleasure  ;  the  other  unknown  varies  also,  but  it  is  de- 
pendent upon  the  values  of  the  first.  In  2  x  —  y  =  4,  if  x  =  0, 
y  =  —  4 ;  if  x  =  2,  y  =  0 ;  if  x  =  3,  y  =  2,  etc. 

But  if  besides  2  x  —  y  =  4,  we  have  a  second  relation  between 
the  variables,  wholly  independent  of  the  first  relation,  as 
2  x  +  y  =  12,  then  we  have  what  is  called  a  system  of  linear 
equations  in  which  x  and  y  are  subjected  to  two  conditions 
which  restrict  x  and  y  to  a  single  value  each. 

The  solution  of  equations  by  graphs  is  not  always  easily  ac- 
complished and  may  not  be  accurate. 

To  solve  the  system  without  graphs  we  must  reduce  the  two 
equations  in  two  unknowns  to  one  equation  in  one  unknown. 
This  process  is  called  elimination. 

The  elimination  may  be  accomplished  in  either  of  two  ways : 
I.  addition,  or  subtraction,  or  II.  substitution. 

SOLUTION   BY   ADDITION   OR  SUBTRACTION 

152.  I.     Solve  x  +    y  =  7.  (1) 

2  x  -  3  y  =  4.  (2) 

Multiply  (1)  by  2,  to  make  the  coefficients  of  x  alike  in  both  equations. 
2x  +  2y  =  U.  (3) 

2x-3y  =4  (2) 

Subtract,  5  y  =  10 

J/  =  2. 
Substitute  2  for  y  in  (1),  x  +  2  =  7. 

x  =  5.     Ans.  5,  2. 
126 


SIMULTANEOUS  LINEAR  EQUATIONS  127 

Check:  Substituting  5  for  x  and  2  for  y  in  (1),  and  also  in  (2),  we 
obtain : 

5  +  2  =  7,  7  =  7. 
10  _  6  =  4T  4  =  4. 

II.     Solve  2x-3y  =  ±.  (1) 

7  a  +  4  //  =  -  15.  (2) 

Multiply  (1)  by  4,  8  x  -  12  y  =  16  (8) 

Multiply  (2)  by  3,  21  x  +  12  y  -  -  45  (4) 

Add,  29  x  =  -  29 

x  =-  1. 

Substitute  -  1  forx in  (1),  —  2  -  3  y  =  4. 
-  3  y  =  6. 

2/  =  -2.     Ans.  -1,-2. 
Check :  Substituting  in  (1), -2  +  6  =  4;  4=4. 

Substituting  in  (2),  -  7  -  8  =  -  15  ;  -  15  =  -  15. 

Notice  that  the  coefficients  of  x  or  of  y  in  the  two  equations  may  be 
made  alike  in  absolute  value  by  multiplying  both  sides  of  one  equation  or 
of  each  equation  by  some  appropriate  number.  Then  either  add  or  sub- 
tract to  eliminate  one  letter.  By  substitution,  find  the  value  of  the  other 
letter. 

EXERCISES 
153.    Solve  and  check  : 

1.  2x-3y  =  -l,  7.   3«-10.s  =  32, 
5  x  +  2  y  =  45.  6 1  -  20  s  =  64. 

2.  7x  +  2y  =  l,  8.    2m  =  3 w, 

4  x  —  y  =  7.  9  m  —  17  n  =  —  7. 

3.  a  =  2  y,  9.   3  a  +  6  b  =  7, 

3  a;  +  7  y  =  130.  6a  +  12  6  =  15. 

4.  6 a -3.?/ =  3,  10.    I0a+3y  =  159, 
x  +  y  =  3.  :;..-+  10//  =  1 66. 

5.  3»-2t/  =  l,  11.   3x-52/  =  23, 
4x  +  3/y  =  2|.  7x  +  //  =  -35. 

6.  5a  +  76  =  14,  12.    2x  +  5y  =  49, 
3a_4&  =  _8.  3a-2y  =  -50. 


128  ELEMENTARY   ALGEBRA 

SOLUTION  BY  SUBSTITUTION 

154.   Solve  2x-3y  =  27.  (1) 

5*+2rf«l.  (2) 

From  (1),  2x  =  27+Sy. 

K=W±M.  (3) 

2  v  J 

Substitute  in  (2) ,    135  +  16y  +  2  y  =  1.  (4) 

Multiply  both  sides  by  2, 

135  +  15y  +  4?/  =  2.  (6) 

Whence  19  ?/  =  -  133. 

y  =  -l. 

[     3.    Am.  3,-7. 


Substitute  in  (3), 

X  — 

2 

Check:  In  (1), 

6  +  21  =27;  27  =  27. 

In  (2), 

15-14  =  1;       1  =  1. 

Notice  that  the  value  of  x  was  found  in  terms  of  y  in  one  equation  and 
substituted  in  the  other  ;  the  resulting  equation  was  solved,  and  the  value 
of  the  other  letter  found  by  substitution. 

Either  letter  may  be  eliminated,  its  value  in  terms  of  the  other  letter 
being  found  in  either  equation  and  substituted  in  the  other  equation. 

EXERCISES 

155.  Solve,  by  substitution,  and  check.  Which  two  equa- 
tions are  inconsistent  (§  146,  Ex.  15)  ?  Which  are  not  inde- 
pendent ? 

1.  4  x  +  5  y  =  10,  5.   x  +  21  y  =  2, 
7x  +  3y  =  6.  27 y  4- 2a  =  19. 

2.  3x  +  2y  =  60,  6.    6p  +  5q  =  2, 
2x  +  3y  =  60.  p  +  3q  =  9. 

3.  4rx  —  6y  =  —  96,  7.   8m  +  5w=-  1, 
10o;+32/  =  120.  4ra  =  10w  +  7. 

4.  3  x  +  5  y  =  50,  8.   fc  =  5 1  +  3, 
x  -  7  y  =  8.  I  =  2  &  -  24. 


SIMULTANEOUS  LINEAR  EQUATIONS  129 


9.   4r  +  3s  =  5, 

11. 

a  +  y  =  7, 

4  s  -  3  r  =  1\. 

2z  +  2?/  =  14. 

10.    a  -26  =  2, 

12. 

3  a;  —  4 1/  =  5, 

2  &  -  6  a  =  3. 

6a;-8y  =  7. 

EQUATIONS 

CONTAINING 

FRACTIONS 

156.   Solve 

m 

+3 

=  4. 

(1) 

m 
6 

n 

~2: 

=  — 

3. 

(2) 

Multiply  (1)  by  3, 

m  • 

3n 
4 

=  12 

(3) 

Multiply  (2)  by  6, 

m  - 

-  8  n : 

=  -18. 

(4) 

Subtract  (4)  from  3, 

3«  +  ^ 
4 

=  30. 

(5) 

Multiply  (5)  by  4, 

12n+3n 

=  120. 

15  n 

=  120. 

n 

=  8. 

Substitute  8  for  n  in  (4), 

m 

-24 

=  - : 

18. 

m  : 

=  6. 

vlras.  m  =  6,  n  =  8. 

Check  this  answer  by  substitution  in 

(1)< 

md  (2). 

EXERCISES 

157.    Solve,  and  check 

-      a      b     4 

3     2~3' 

4. 

aj      »__  1 

3      4-l2' 

a     b      7 
2     3     6" 

a;     3y      1 
2      16      2* 

2-    ?-£  =  <>, 

8     9       ' 

5. 

™  +  ^  =  3, 
7      6        ' 

abb 
3  ~  4  "  12' 

3m             iK 
n  =  15. 

2 

3.    »-I«|a 

2     2' 

6. 

5        5 

*        3       3 

O         o 

130  ELEMENTARY   ALGEBRA 

7.        ^  +  5*  =  9,  10.    ^p-^=0, 

s  +  9      r-2_A  2a- 7      13  -  b 


10  3 


=  0.  ""  '  -    "~     =10. 


wL+2_v--3  =  _2  tl     5  +  3a     5y-2_     2 

9  2  '  7  4 

^-^=-1-  6z  +  8<,=108. 

:c+j/_+2_3  a  +  6^      5 

x-y  +  2       '  '    b-a         9' 

»-y-2  =  l.  3a-36  +  11  =  20f 

x  +  y  — 2      5  7  7 

13.  a  =  2  y, 

x  +  y      x  —  y  _3x  —  2y     3y  —  x     31 
5  5  ; 3  2  10* 

14.  *  =  2, 

y    3' 

,     5 a;  +  6 y     4  a;  —  3y  __     _4 
10  3~~     _2/      3' 


LITERAL  SIMULTANEOUS  EQUATIONS 

158.    Solve  ax  +  y=b,  (1) 

ex  —  dy  =  e.  (2) 

Multiply  (1)  by  d,  a<fa  +  ay  =  &a\  (3) 

ex  —  dy  =  e.  (2) 

Add  (2)  and  (3),  (ad  +  c)x  =bd+  e.  (4) 

Divide  both  sides  of  (4)  by  (ad  +  c), 

ad  +  c 
By  substituting  this  value  of  a;  in  (1)  or  in  (2)  we  can  find  y. 
But  it  is  easier,  in  this  case,  to  find  y  by  the  same  method  which  gave 
x.     Accordingly, 

Multiply  (1)  by  c,  acx  +  c,y  =  be.  (5) 

Multiply  (2)  by  a,  acx  —  ady  =  ae.  (6) 


SIMULTANEOUS  LINEAR  EQUATIONS  131 

Subtract  (6)  from  (5),       (c  +  ad)y  =  be  —  ae. 

be  —  ae 

V  = • 

c  +  ad 

EXERCISES 
159.    Solve: 

1.  x  +  y  =  a,  10.    m+a?i  =  0, 
x  —  y  =  b.          •  bm  +  n  =  1. 

2.  aa;  +  ty  =  c,  n     2  6x-3a?/  =  c, 
x~y=d-  3bx+2ay  =  d. 

3.  ex  —  dy  =  m, 

dx  +  cy  =  7i.  12.    - •  +  ^  =  c, 

a     6 

4.  »  =  ay,  x_y  =  d 
bx  +  dy  =  bcd.  a      &~ 

5.  mx+?i?/  =  l,  ^ 

c.r  —  dy  =  l.  13-      "^I8*** 

6.  3  x  —  2y  =  m  +  w,  v     x 
2x  —  3y  =  m—  n.  b     a 

7.  ar— jf  =s  lj 

(a  +  &)x-(a-%  =  0.         14.    ^  +  ^  =  ™> 

8.  (a  -  6)x  -  y  =  0,  r+s=n 
re  +  (a  —  &)?/  =  0.  6      a 

9.  r  +  ds  =  3, 
dr  — -  s  =  3  d. 

15.  Solve  for  R:     0=2  wR. 

16.  Solve  for  a  and  6  :     ^4.  =  -- . 

17.  Solve  for  a  and  h :     A  =  \{a  +  b)h. 

18.  Solve  for  r  and  t :     d  =  rt. 

19.  Solve  for  3f :     F  =  Ma. 


132  ELEMENTARY   ALGEBRA 


20.  Solve  for  ft,  I,  and  n  :     #  =  -  (tt  +  l). 

Z 

21.  Solve  for  r  and  « :     ^  =  P  +  Prt. 

22.  Solve  for  d  and  n :     l  =  a  +  (n  —  l)d. 

23.  Solve  for  2^:     C  =  f(i^- 32). 

24.  Solve  for  s  and  a :     —  =  — -  • 

160.   Solve  ?  +  §  =  20,  .     (1) 

x     y 

---  =  17.  (2) 

x     y 

If  we  multiply  both  sides  of  each  equation  by  xy,  the  resulting 
equations  are  much  more  difficult  to  solve.  Hence  do  not  multiply  by  xy, 
but  eliminate  the  fraction  containing  either  x  or  y. 

Check  : 

(1) 

(2) 


Multiply  (1)  by  3, 

and 

(2)  by  2, 

-  +  -  =  60 
x     y 

(3) 

CTiec*  : 
14  +  6  =  20. 
20  =  20. 

«-*=34 
x     y 

(4) 

21  -  4  =  17. 

Subtract, 

1^  =  26 

2/  =  i 

17  =  17. 

In  (1)  substitute  \ 

for  y 

,  -  +  6  =  20, 

X 

X 

2  =  14x, 

x=*. 

^4ws. 

M- 

EXERCISES 

161.    Solve  and  check : 

L    M— 27,  2.  |-*-|, 

»    y  x    y    2 

15_3=_39  1  +  3  =  5. 

a;      y  ay 


SIMULTANEOUS   LINEAR  EQUATIONS  133 


5. 


5+ 

X 

3__ 

y 

5, 

8_ 

^  =  31 

X 

2/ 

3 

2x 

5 
3y 

li» 

5 
3a; 

+h= 

4|. 

a; 

1 

-  =  a, 

1_ 

.1-6. 

X 

y 

a  i  1      k 
-  +  -  =  o, 
x     y 

c 

1-* 

7. 

2      m 

=  a, 

99       y 

?  +  *-& 

99       1/ 

8. 

aa;      oy 

6a;     ay 

9. 

6       5      - 

ax     &y 

ax     by 

0. 

( =  ?n-  +  n, 

na;     my 

—  H —  =  m2  +■  n2. 
98      y 

x     y 


THREE  SIMULTANEOUS   LINEAR  EQUATIONS 

162.  If  the  system  consists  of  three  independent  equations 
containing  three  unknowns,  one  of  the  unknowns  must  be 
eliminated  between  two  pairs  of  the  equations.  The  resulting 
equations  may  then  be  solved  and  the  third  unknown  found  by 
substitution. 


Solve                                 x  +  y  —  z  =  0, 

(1) 

2x-y  +  3z  =  9, 

(2) 

3z  +  2y  +  z  =  10. 

(3) 

Eliminate  y  between  (1)  and  (2), 

x  +  y  —    z  =  0 

(1) 

2x  -y  +  32  =  9 

(2) 

Add,                                            3  x      +2^  =  9 

(4) 

Eliminate  y  between  (2)  and  (3),  by  multiplying 

(2)  by  2, 

4x-2y+6«  =  18 

(5) 

3x  +  2y+    0  =  10 

(3) 

Add,                                           7  x  +           7  z  =  28 

(6) 

134  ELEMENTARY   ALGEBRA 

Divide  by  7,  z  +  z  =  4.  (7) 

Eliminate  z  between  (4)  and  (7),  by  multiplying  (7)  by  2, 

2x  +  2z  =      8  (8) 

3x  +  2g  =       9  (4) 

Subtract,  —  x  =  —  1. 

x  =  l. 
In  (7)  substitute  1  for  x,  1  +  z  —  4  ;  z  =  3. 

In  (1)  substitute  1  for  x,  and  3  iorz,   1+y  — 3  =  0;  y  =  2. 

Jtis.  1,  2,  3. 
Be  sure  that  after  eliminating  one  unknown  between  two  pairs  of  the 
equations,  you  have  two  equations  containing  the  same  two  unknowns. 

Should  there  be  more  than  three  unknowns,  reduce  the  system  to  a 
system  containing  one  less  equation  and  one  less  unknown.  Continue 
this  process  until  you  have  one  equation  with  one  unknown. 

EXERCISES 
163.    Solve  and  check : 

1.  2x-3y  +  4z  =  20,  7.  2r-3s  =  5, 
3x+y-z  =  $,  3r  +  t  =  10, 
5x-*/  +  3z  =  28.  s-2t  =  5. 

2.  x  +  y  +  z  =  2,  8.  3a +  56  =  95, 
x—  ?/  +  z  =  16,  a  —  2  c  =  —  9, 
x  +  y  —  z  =  —  2.  7c  —  46  =  44. 

3.  2x-5y-4z= -4,  1  1  1_ fl 
x+y  +  z  =  0,  9-  a+b+c~ ' 
4a-3y-2z=-4.  i      1  +  1=2 

4.  4a -36  + 3c  =  20,  a~~6  c  ' 
-7a  +  26  +  6c  =  5,  1  ,  1  _  1  ^  Q 
8a  — 6  +  2c  =  45.  a     6     c 

5.  2a-36  +  4c  =  2,  l      \_ 

3a -3  6- 15  =  0,  10,    x  +  y~~    ' 

7a_4c-31  =  0.  1  +  1=_2 

6.  4  m  — 5  w  =  22,  *     z 

6  m  — 5p  =  — 12,  1  ,  l_o 

lOn  +  3p  =  -2.  y      z~    ' 


SIMULTANEOUS  LINEAR  EQUATIONS  135 

„     2  S,4_19  12.   a  +  6  +  c+d  =  l.l, 

a  b      c  a  —  o  —  c-\- a=—.6, 

5  6     7_  _28  a  +  2&  +  3c-4d  =  .8, 

a  6      c  a  +  6-4c+3d=-.8. 

1-1  +  5-  -43. 

a      6      c 

PROBLEMS 

164.  1.  The  sum  of  two  numbers  is  78;  their  difference  is 
4.     Find  the  numbers. 

2.  The  sum  of  two  numbers  is  76;  their  difference  is  26. 
Find  the  numbers. 

3.  If  the  difference  of  two  numbers  is  12  and  their  sum  is 
45,  what  are  the  numbers  ? 

4.  There  are  two  numbers  whose  sum  is  25.  If  the  larger 
be  subtracted  from  the  smaller,  the  remainder  is  —  6.  Find 
the  numbers. 

5.  Twice  a  certain  number  plus  another  number  is  35.  The 
difference  between  the  two  numbers  is  1.  What  are  the 
numbers  ? 

6.  The  sum  of  two  numbers  is  72.  If  the  second  be  sub- 
tracted from  twice  the  first,  the  remainder  is  3.  Find  the 
numbers. 

7.  The  difference  between  two  numbers  is  10 ;  their  sum 
is  —  98.     What  are  the  numbers  ? 

8.  The  sum  of  two  numbers  is  26;  their  difference  is  2. 
Find  them. 

9.  A  father  earns  $45  more  a  month  than  his  son. 
Together  they  earn  $125  a  month.  What  is  the  monthly 
salary  of  each  ? 

10.  A  real  estate  dealer  purchases  two  houses  for  $11,250; 
one  house  is  worth  $250  more  than  the  other.  Find  the 
value  of  each. 


136  ELEMENTARY   ALGEBRA 

11.  A  mother  is  21  years  older  than  her  daughter ;  their 
combined  ages  are  71  years.     Ascertain  the  age  of  each. 

12.  Five  pounds  of  sugar  and  three  dozen  of  eggs  cost  $  1.20 ; 
seven  pounds  of  sugar  and  five  dozen  of  eggs  cost  S1.92. 
What  will  one  dozen  of  eggs  and  one  pound  of  sugar  cost  ? 

13.  I  bought  10  yards  of  silk  and  5  yards  of  cambric  for 
$14.  If  I  had  bought  2  yards  less  of  silk  and  3  yards  more 
of  cambric,  my  bill  would  have  been  $  1.60  less.  What  is  the 
price  of  each  per  yard  ? 

14.  Find  two  numbers  such  that  the  greater  exceeds  the 
less  by  11,  and  one  fourth  the  less  plus  one  third  the  greater 
is  13. 

15.  Find  the  fraction  in  which  if  1  be  added  to  the 
numerator  its  value  is  1,  but  if  6  be  added  to  the  denominator 
the  value  is  \. 

16.  The  sum  of  three  numbers  is  29  ;  twice  the  first  plus  3 
equals  the  second ;  twice  the  third  minus  1  equals  the  second. 
Find  the  numbers. 

17.  A  man  hired  3  boys  for  a  day  for  $4.25.  The  second 
boy  received  half  again  as  much  as  the  first,  and  the  third 
received  25^  more  than  the  second.  How  much  did  each 
receive  ? 

18.  Find  two  numbers  such  that  three  times  the  greater 
exceeds  twice  the  less  by  61,  and  twice  the  greater  exceeds  3 
times  the  less  by  4. 

19.  A  and  B  have  equal  sums  of  money.  A  spent  half  of 
his  and  B  earns  $  10 ;  then  B  has  3  times  as  much  as  A.  How 
much  had  each  at  first  ? 

20.  If  A  gives  B  $5,  A  will  have  \  as  much  as  B;  but  if  B 
gives  A  $25,  B  will  have  the  same  amount  as  A.  How  much 
has  each  ? 


SIMULTANEOUS  LINEAR  EQUATIONS  137 

21.  A  can  do  a  piece  of  work  in  3  days,  B  can  do  the  same 
work  in  4  days.     In  what  time  will  both  together  do  the  work? 

22.  A  and  B  together  do  some  work  in  4  days,  A  and  C 
together  in  5  days,  B  and  C  in  10  days.  How  long  would  it 
take  each  alone  ? 

23.  A  company  of  boys  bought  a  boat.  If  there  had  been  5 
boys  more,  each  would  have  paid  $  1  less ;  but  if  there  had 
been  3  boys  fewer,  each  would  have  paid  $1  more.  How 
many  boys  were  there  ?  What  did  each  pay  ?  What  did  the 
boat  cost  ? 

24.  A  two-digit  number  has  the  sum  of  its  digits  equal  to  9. 
If  9  be  added  to  the  number,  the  digits  will  be  interchanged. 
Find  the  number. 

Let  t  —  the  digit  in  tens'  place,  u  =  the  digit  in  units'  place.  Then 
lOt  +  u  =  the  number. 

25.  The  tens'  digit  is  twice  the  units'  digit.  If  36  be 
subtracted  from  the  number,  the  digits  will  be  reversed.  What 
is  the  number  ? 

26.  The  sum  of  the  digits  of  a  three-digit  number  is  11. 
The  units'  digit  is  twice  the  tens'.  If  the  digits  are  reversed, 
the  number  will  be  396  more  than  it  is.     Find  the  number. 

27.  Three  cities  are  located  at  the  vertices  of  a  triangle. 
The  distance  from  A  to  B  by  way  of  C  is  150  miles ;  from  C 
to  A  by  way  of  B  is  140  miles ;  from  B  to  C  by  way  of  A  is 
110  miles.     How  far  apart  are  the  cities  ? 

28.  If  a  rectangle  were  1  inch  longer  and  2  inches  narrower, 
it  would  contain  20  square  inches  less;  if  it  were  3  inches 
wider,  it  would  become  a  square.     Find  its  dimensions. 

29.  A  train  maintains  a  uniform  speed  for  a  certain  dis- 
tance. If  the  speed  had  been  10  miles  an  hour  less,  the  time 
would  have  been  2\  hours  more;  if  the  speed  had  been  10 
miles  an  hour  more,  the  time  would  have  been  1|  hours  less. 
Find  time,  rate,  distance. 


138  ELEMENTARY   ALGEBRA 

30.  A  man  invests  a  certain  sum  of  money  at  4  <f0  and  a 
second  sura  at  3^  </0  •  His  income  from  the  two  investments  is 
$445.  Had  the  first  sum  been  invested  at  5%,  his  yearly- 
income  would  have  been  $50  more.  How  much  had  he  in 
each  investment  ? 

31.  B's  yearly  income  is  f  of  A's,  and  A  spends  $500  per 
annum  more  than  B.  At  the  end  of  3  years  A  has  saved 
$  4500  and  B  $  2250.     What  is  the  yearly  income  of  each  ? 

32.  A  person  had  $  1750  invested  so  as  to  bring  in  an  annual 
income  of  $  77.  Part  was  bearing  4%  interest,  the  rest  5%. 
How  much  was  invested  at  4  %  ? 

33.  A  merchant  mixes  two  qualities  of  tea  in  the  ratio  of 
two  parts  of  the  cheaper  to  one  part  of  the  dearer,  and,  by  sell- 
ing the  mixture  at  60  ^  per  pound,  he  gains  a  profit  of  25  °f0  ; 
on  mixing  them  in  the  ratio  of  three  parts  of  the  cheaper  to 
two  parts  of  the  dearer,  and  selling  the  mixture  at  65^  per 
pound,  he  gains  the  same  rate  of  profit.  Find  the  prices  at 
which  he  bought. 

THE  BEGINNINGS  OF  ALGEBRA 

165.  The  earliest  traces  of  algebra  are  found  in  a  very  old  Egyptian 
papyrus,  written  by  Ahmes  about  1700  or  2000  b.c.  This  papyrus  is  a 
copy  of  a  still  older  document.  The  Ahmes  papyrus  was  found  in  1868 
and  is  the  most  important  source  of  our  knowledge  of  ancient  Egyptian 
mathematics.  It  contains  interesting  matters  on  fractions,  geometry, 
algebra,  and  even  the  rudiments  of  trigonometry.  It  gives  solutions  of 
linear  equations.  It  might  be  expected  that  the  symbols  used  3000  years 
ago  would  be  as  different  from  ours  as  their  language  and  speech  were 
different  from  ours.  To  give  an  idea  of  the  writing  in  the  Ahmes  papy- 
rus, we  reproduce  one  equation  and  place  beneath  it  the  English  transla- 
tion and  the  equation  written  in  modern  symbols. 

nnn 
mini 

Heap,  its  f ,  its  \,  its  1.         its  whole,  it  makes         :  37 

That  is,  x  (|     +     f     +     }      +      1)  =  37. 


SIMULTANEOUS  LINEAR  EQUATIONS  139 

Notice  particularly  that   <rx>  means  |,    /  means  £,  <ZZ>    is  the 

symbol  to  indicate  that  7  is  the  denominator  of  a  unit  fraction,  p  means 
10. 

No  further  progress  was  made  in  algebra  until  the  time  of  the  Greek 
Diophantus  (about  300  b.c),  who  was  very  skillful  in  solving  problems. 
He  had  a  symbol  of  his  own  for  the  unknown  x  and  for  x2.  He  had 
also  a  symbol  to  express  subtraction  and  equality.  Juxtaposition,  which 
with  us  represents  multiplication,  meant  addition  with  him.  A  most 
momentous  impulse  to  the  development  of  algebra  was  given  by  the 
Hindus,  after  the  fifth  century  a.d.  They  displayed  wonderful  skill  in 
developing  arithmetic  and  algebra.  The  Hindu  and  Greek  knowledge  of 
algebra  was  later  transmitted  to  the  Arabs  who,  in  turn,  became  the 
teachers  of  European  nations  at  the  close  of  the  dark  ages.  The  early 
European  writers  on  algebra  used  hardly  any  symbols  ;  everything  except 
arithmetical  numbers  was  fully  written  out  in  words.  In  the  sixteenth 
century  the  modern  symbols  began  to  be  invented.  These  symbols  con- 
stitute a  sort  of  shorthand  method  of  indicating  algebraic  relations  and 
processes,  whereby  the  power  and  usefulness  of  algebra  in  the  solution  of 
problems  is  very  greatly  increased.  How  important  they  are  can  be 
realized  when  the  attempt  is  made  to  solve  difficult  problems  without  their 
aid.  The  sixteenth-century  symbols  are  more  clumsy  and  less  convenient 
than  the  modern.  The  great  French  algebraist,  Vieta,  wrote  in  1595  the 
polynomial 

2-16  x2  +  20  x*  -  8  x6  +  x8,  thus :  2  -  16  Q  +  20  QQ  -  8  CC  +  QCC. 

Here  Q  (quadratus)  stands  for  x2,  C  (cubus)  stands  for  x3.     The  Portu- 
guese mathematician,  Pedro  Nunez,  in  1567,  wrote 

y/VT&6  +  27  --%/Vfm-  27  thus  : 

R.  V.  cu.  R.  756  •  p27  •  m  •  R.  V.  cu.  R.  756  .  m  •  27. 

Here  R.  V.  cu.  means  radix  universalis  cubica  or  general  cube  root, 
p  means  plus,  m  means  minus.  The  notations  of  Vieta  and  Descartes 
will  be  illustrated  by  other  examples  which  we  place  under  their  portraits 
opposite  pages  110  and  140.  The  notation  of  Descartes  was  nearly  the 
same  as  the  modern  notation. 

The  sign  of  equality  ( = )  was  first  used  for  this  purpose  by  an  English 
writer  on  arithmetic  and  algebra,  Robert  Recorde.  The  following  photo- 
graph of  part  of  a  page  in  his  algebra,  entitled  the  Whetstone  of  Witte, 


140  ELEMENTARY   ALGEBRA 

published  in  London  in  1557.  shows  the  place  in  that  book  where  the 
sign  =  is  first  introduced. 


'  lftofobett,foj  eaffc  atteratfo  of  e  $utim.%  Ml  p^o> 
pounbe  a  fetoc  crapleg, bicau fc  the  extraction  of  tbetr 
rootcs,maie  tijc  mo;c  aptlp  bee  to;ougbte.  3n0  to  a* 
uoioc  tbc  tcbioufe  repetition  of  tbcfc  IucojOcs :  isc* 
quallc  to :  31  Uitll  ferte  as  3;  toe  often  in  U)oo;fec  tofc,a 
pat  re  of  paraUelcs,o;  Ocmotue  itnes  of  one  leitgtbe, 
tl)us:==^===,btcaufc  noc.2.  tbpngcs,can  be  moan? 
equalle.  3no nolo  marlw  tbcfc  nonibcrs. 


J. 

4.     !  9.3^—1 — 1 9  2.f «~ i'oj. — I — ioSf 1 9  S£ 

1 8.^—f— 2  4-f .=-  8.5-.— f—  2.=£. 

545" 125^— 40^-- r— 480? — 9.5- 

n  tbc  firftc  there  appearetb.  2  ♦  nmnfarrf ,  tbat  w 

I4^f/ 


From  Recorde's  Whetstone  of  Witie,  1557. 
Fig.  27 


REN£   DESCARTES 


Descartes'  Notation 
z8  *  +  px  +  qxO; 


Modern  Notation 
x3  +  px  +  q  =  0. 


Descartes  used  oo  as  the  sign  of  equality.     The  *  after  x3  signified  that 
the  next  lower  power  of  x  was  missing  from  the  expression. 


CHAPTER   XIII 

QUADRATIC   EQUATIONS 

166.  The  equation  x2  +  ax  +  b  =  0 

is  called  a  complete  quadratic  equation.  It  is  a  quadratic  equa- 
tion because  the  highest  power  of  the  unknown  x  is  the 
second ;  it  is  complete  because  it  contains  also  a  term  involv- 
ing the  unknown  x  to  the  first  power  and  a  term  b  (often 
called  the  absolute  term)  which  is  free  from  x. 
Quadratic  equations  of  the  forms 

x2  =  b, 

x2  -+-  ax  =  0, 
are  called  incomplete  quadratic  equations,  because   either   the 
term  involving  the  unknown  x  to  the  first  power  or  the  abso- 
lute term  b  is  absent. 

INCOMPLETE    QUADRATIC   EQUATIONS 

167.  Equations  of  the  form  x2  +  ax  =  0  can  be  solved  by 
the  method  of  "  completing  the  square,"  to  be  explained  later. 
Much  easier  is  the  solution  by  the  method  of  factoring,  treated 
in  a  previous  chapter. 

Factoring,  we  obtain  x{x  +  a)  =  0. 

Making  the  first  factor  equal  to  zero,  x  =  0. 

Making  the  second  factor  equal  to  zero,        x  +  a  =  0. 

x  =—  a. 

Very  simple  is  the  solution  of  x2  =  b.  For  greater  conven- 
ience, write  it  x2  =  c2. 

Extracting  the  square  root  of  both  sides,  x  =  ±  c. 

Since  the  square  root  of  both  sides  has  been  extracted,  it  might  be 
claimed  that  the  sign  ±  should  be  written  on  both  sides,  giving 

±  x  =  ±  c. 
141 


142  ELEMENTARY   ALGEBRA 

But  this  result  is  the  same  as  when  we  write        x  =±c. 
For  the  equation  ±x  =±  c  means  here 

(1)  +x  =  +  c.  (3)   +x  =  -c. 

(2)  -x=-c.  (4)  -a;  =  +c. 

Of  these  four  sets,  the  first  two  are  the  same,  and  the  last  two  are  the 
same.     Hence,  x  =  ±  c  gives  all  the  values  of  x. 

EXERCISES 

168.   1.    Solve  ^--3  =  ^+5. 
2  2 

Transposing  and  combining, 

4a;2  =  8. 
Hence  x2  =  2, 

and  x  —  ±  V2. 

Check:  ®ll-S  =  ?+  5,  or  6  =  6. 

2  2 

Solve  the  following : 

2.   4  x2  -  4  =  ±£  -  5  x2.  a?  +  2  __  a?  — 2  g 

3      *  +  *-3.  3  2 

'2a;      a:  „     ^-3  ,  3x2-4 


4.  i  +  A  =  2. 

a,*2      x2 

5.  4i/2  =  32/2+ir. 

6.  7  a;2 -63  =  7.  14.   3a; --^-  =  0. 


11.  tLJZ^.  +  "  f  ~  •*  =  13. 
4  2 

12.  2  x2 +  5.78  a;  =  0. 

13.  7  a;2  =  10.5  a;. 


7.    llz2-43  =  5z2-7. 


3a; 

.  —  -6-.4a;  =  0. 


s  2y2-8=2 

5  16.    .4*-^  =  0. 

9    3x2  +  15  =  3  * 

13  17.    .01  x2  -  .08  a;  =  0. 

COMPLETE  QUADRATIC  EQUATIONS 

169.   In  §  131  we  solved  complete  quadratic  equations  by 
the  method  of  factoring.     That  method  is  the  best  when  the 


QUADRATIC  EQUATIONS  143 

factors  can  be  found  by  inspection.  But  there  are  equations, 
like  cc2-f-3a;-f-l  =  0,  which  cannot  be  solved  by  any  method 
of  factoring  thus  far  studied.  Let  us  consider  another  method 
of  solving  complete  quadratic  equations. 

Of  what  expression  is  the  trinomial  a2  -f  2  ab  +  b2  a  perfect 
square  ?  Of  what  is  x2  +  2  bx  +  b2  a  perfect  square  ?  What 
must  be  added  to  x2  +  2  bx,  to  make  the  resulting  trinomial  a 
perfect  square  ?  How  can  b2  be  found  from  x2  +  2  bx  ?  The 
process  of  finding  this  third  term,  b2,  when  the  two  terms 
x2  -f-  2  bx  are  given  is  called  completing  the  square. 

Rule.  To  complete  the  square  of  x2  +  2  6a;,  ao?d  to  this  the 
square  of  half  the  coefficient  of  x. 

ORAL   EXERCISES 

170.  Complete  the  square  of  the  following : 

1.  x2  +  2x.         3.   x2  -  4 x.         5.    z2  +  3z.          7.  y2  +  16 y. 

2.  a2 -f  6  a.       (  4.   y2  —  Sy.  6.    aj2  +  5cc.         8.   z2  +  llz. 

SOLUTION  OF  QUADRATIC  EQUATIONS  BY  COMPLETING  THE 

SQUARE 

171.  1.    Solve  the  equation  x2  -f  6x  —  7  =  0. 

(1)  Transpose  the  term  —7  which  does  not 

involve  x  or  a;2.  x2  +  6  x  —  7  =  0. 

(2)  Complete  the  square  in  the  left  member, 

by  adding  32  to  both  sides.  x2  +  6  x  +  9  =  16. 

(3)  Take  the  square  root  of  both  sides.  x  +  3  =  ±  4. 

(4)  Transpose  the  3  to  the  right  side.  x  =  ±  4  —  3. 
Simplify.  x  =  +  1  or  —  7. 
Checking  in  the  original  equation,  x2  +  6  x  —  7  =  0. 

The  value  x  =  1  gives :  The  value  x  =  —  7  gives  : 

l  +  6(l)-7=0.  (_7)S+6(-7)  -  7=0. 
1  +  6  -  7  =  0.  49  -  42  -  7  =  0. 

Hence  1  is  a  root.  —  7  is  a  root. 

The  reason  for  substituting  the  values  of  x  in  the  original  equation, 
rather  than  in  some  equation  derived  from  it,  is  evident.  An  error  may 
have  been  committed  in  getting  the  derived  equation,  but  this  derived 


144  ELEMENTARY   ALGEBRA 

equation  may  be  correctly  solved.     Substituting  the  values  of  x  in  the 
derived  equation  would  not  reveal  the  error. 

2.    Solve  9o2-5=-4a. 

(1)  Transpose  —  4  x  to  the  left  side  and  —  5  to  the  right, 

9  x2  +  4  x  =  6. 

(2)  Divide  both  sides  by  9,  the  coefficient  of  x2, 

X2+$X  =  l 

(3)  Complete  the  square  by  adding  (f)2  to  both  sides, 

(4)  Find  the  square  root  of  both  sides,  x  +  $  =  ±  £. 

(5)  Transposing  and  combining,  x  =  ±  $  —  f . 

*  =  *-*  =  $• 
x=-$-f  =-1. 
Checking  in  the  original  equation,  9  x2  —  5  =—  4  x. 

The  value  x  as  —  1  gives :  The  value  x  =  \  gives : 
9(-l)2-5=-4(-l).  9(f)2 -5  =-4(|). 

9-5  =  4.  ^_6=_^i. 

4  =  4.  _¥=_^i. 

Hence  —  1  is  a  root.  Hence  f  is  a  root. 

172.  To  solve  complete  quadratic  equations : 

(1)  Transpose  the  terms  in  x?  and  in  x  to  the  left  side,  and  the 
terms  free  of  x  to  the  right. 

(2)  Divide  both  sides  by  the  coefficient  of  x2. 

(3)  Complete  the  square  by  adding  to  each  side  the  square  of 
half  the  coefficient  of  x. 

(4)  When  possible,  simplify  the  right  side  by  combining  its  terms. 

(5)  Extract  the  square  root  of  both  sides,  and  use  both  signs 
±  on  the  right. 

(6)  Find  the  two  values  of  x. 

EXERCISES 

173.  Solve  the  following  quadratic  equations  by  completing 
the  square,  and  check : 

1.  x2  +  6a -16  =  0.  4.   25a2  +  20a  =  21. 

2.  z2  +  10a:  =  24.  5.    36a2  +  24a  =  45. 

3.  4a2  +  12a  =  27.  6.    9y2  =  6y  +  63. 


QUADRATIC   EQUATIONS  145 

7.    Solve  9s2  4- 36x- 2  =  0. 

(1)  Transpose  -  2,  9  x2  +  36  x  =  2. 

(2)  Divide  both  sides  by  9,  x2  +  4  x  =  - . 

(3)  Complete  the  square  by  adding  4  to  both  sides, 

x2  +  4x  +  4  =  38 

(4)  Find  the  square  root  of  both  sides, 

3 

(5)  Solve  for  x,  x=-2±^§. 

The  value  of  the  radical  V38  cannot  be  computed  accurately.     Taking 
V38  =  6.16  +,  we  obtain  the  following  approximate  values  for  x  : 
_  +6.16  _  2  _  2  05  _  2  _  Q5 
3 
X  =  -  2.05  -  2  =  -  4.05. 
Except  when  the  approximate  values  are  needed  in  some  computation, 
it  is  customary  to  retain  the  radical  in  the  expression  for  x ;  thus, 

o   ,  V38 

Solve  the  following : 

8.  4?/2  +  20*/  =  15.  20.   4x2-20x  +  5  =  0. 

9.  9 y2 -  12 y  =  10.  21.    10-7x  =  25o2. 

10.  49  x2  -  28  x  =  16.  22.  16  x2  -  5  x  =  4. 

11.  lOz  —  l  =  z2.  23.  3x  +  5  =  9a:2. 

12.  81z2  =  13-54a;.  24.  3x  =  10-16a2. 

13.  21  -  200  x  =  100  x2.  25.  36x2  +  24a  =  5. 

14.  100  y2  =  25  +  50  y.  26.  2/2  +  7y  =  13. 

15.  121  z2  -  22 2  =  8.  27.  13  =  x2  +  14x. 

16.  x2  +  fx  =  l.  28.  21 -5x-2x2  =  0. 

17.  4  x2  -  6  x  =  0.  29.  3  x2  +  8  x  =  2. 

18.  17?/2  +  34?/=0.  30.  5?/2  +  9?/-l  =  0. 

19.  2z2-5z  =  —  1.  31.  4x  — l  =  x2. 


146  ELEMENTARY   ALGEBRA 

32.  2x  +  5  =  3 x2.  38.    x2  +  2 x  +  ff  =  0. 

33.  a2  —  a  +  |  =  0.  39.    4  x2  +  4  x  -  15  =  0. 

34.  16  =  11  x  4-  3 x\  40.   9a;2  -12a:  -5  =  0. 

35.  15 x  =  18-  2 x2.  41.   a;2  -3a:  =  5. 

36.  4a:2  =  6 x  +  1.  42.   3x2-5  =  5x  — x*  +  2. 

37.  4x2  —  5  =  9a:.  43.   6 x2  —  7x  -  5  =  3 x  +  5. 

44.    Solve  x2  -  2  a:  -  31  =  0. 
We  have  x2  —  2  x  +  1  =  32. 

x  -  1  =  ±  V32. 

x  =  1±  V32  =  l  ±4V2. 
The  radical  V32  can  be  simplified  as  follows  (§  111):     As  32  =  2.16, 
and  16  is  a  perfect  square,  we  have  V32  =  V16.2  =  4V2.     Proceeding  in 
the  same  way,  show  that  V50  ss  5V2,  V8aT2  =  2 as V2,  V32o8  =  4 a V2a. 

45.  Simplify  the  following  radicals : 

Vl8,  V8,  Vl2~a2,  V50o^,  V300x*. 

Solve  the  following: 

46.  x2  -  4  x  -  4  =  0.  49.  9  x2  -  12  x  —  46  =  0. 

47.  x2-6x-9  =  10.  50.  16x2-40x  =  175. 

48.  4x2-12x  =  23.  51.  x2-6x-ll  =  0. 


52.   Solve 


2x 


X— 1      x+ 1 


When  x  appears  in  the  equation  in  the  denominators  of  fractions, 
multiply  both  sides  of  the  equation  by  a  number  which  will  remove  the 
denominators.    In  this  case  multiply  by  (x  —  l)(a;  +  1).     We  obtain 

(x-l)(s  +  l)_2x(x-l)(x  +  1) 
x— 1  x + 1 

Dividing  both  terms  of  each  fraction  by  the  factor  common  to  those 
terms,  we  get 

Jfr- — *?(g  +  l)_2x(x-  l)£x-4-*7 
jx. — 1  jt-f-iT 


QUADRATIC  EQUATIONS  147 


Or  x  +  1  =  2x(x-l). 

2x2-3x  =  1. 

x2-fx  =  £. 


-i-*3?- 

■-!* 

4    ' 

Solve  the  following : 

53.     -5—  =  *. 
x-1     2 

58. 

as  —  4     a; 

54                     —  X 

59. 

1     +     *     = 
a;  +  1      a;  —  4 

3 

2x-l 

10' 

__     x  + 1      1 
55.                —     x. 

60. 

2             x 

x-1     2 

x  +  1     x  —  l' 

56.   -20__20  =  1. 

a:  —  1       cc 

61. 

y         2 

y  +  5     y +  2' 

2,1 

57.   y  =  ^y  +  ~. 

62. 

3            z 

z  —  5     z  —  4" 

PROBLEMS 

174.    1.   The  sum  of  two  numbers  is  10,  their  product  is  24. 
Find  the  numbers. 

Let  x  =  one  of  the  numbers,  then 

10  —  x  =  the  other  number. 
x(10  —  x)  =  their  product,  which  must  be  equal  to  24. 
Hence  x(10  —  x)  =24. 

x2-  10x  =  -24. 
x2-10x  +  25  =  1. 

x  =  ±  1  +  5. 
x  =  6  or  4. 
10  -  x  =  4  or  6. 
Hence  the  two  numbers  are  6  and  4. 

2.   The  product   of  two  numbers  is  148;  their  sum  is  24. 
Find  the  numbers. 


148  ELEMENTARY   ALGEBRA 

3.  The  area  of  a  rectangle  is  112  sq.  in. ;  the  sum  of  its 
length  and  breadth  is  23  in.     Find  its  length  and  breadth. 

4.  Find  the  length  and  breadth  of  a  rectangle  whose  area 
is  4  sq.  in.  and  whose  perimeter  is  8T^  in. 

5.  The  perimeter  of  a  window  frame  is  18  ft. ;  the  area  of 
the  window  is  19^  sq.  ft.     Find  the  dimensions  of  the  window. 

6.  If  a  rectangular  table  is  7  ft.  longer  than  wide,  and  it 
contains  36f  sq.  ft.,  what  are  its  dimensions  ? 

Are  both  roots  of  the  quadratic  equation  to  which  the  problem  leads 
applicable  to  the  problem  ? 

7.  A  tinner  wants  to  make  a  square  box  2  inches  deep, 
with  a  capacity  of  72  cubic  inches.     From  each  corner  of  a 

square  sheet  of  tin  a  2-inch  square  is  cut ; 

the  four  rectangular  strips  thus  made  are 

turned  up.     What  must  be  the  dimensions 

of  the  sheet  of  tin  ? 

8.   The  longer  side  of  a  rectangular  box 

exceeds  the  shorter  side  2  in.     The  box  is 

3  in.  deep  and  has  a  capacity  of  105  cu.  in. 

If  made  out  of  a  single  piece  of  cardboard, 
what  were  the  dimensions  of  the  cardboard,  supposing  that 
there  is  no  other  waste  than  the  3-inch  squares  cut  from  each 
corner  ? 

9.  A  flower  bed  is  12'  x  15'.  How  wide  a  walk  must  sur- 
round the  bed,  to  increase  the  total  area  by  160  sq.  ft.  ? 

10.  The  area  of  a  rectangle  exceeds  that  of  a  square  by 
24  sq.  in.  If  the  side  of  the  square  is  equal  to  half  a  shorter 
side  of  the  rectangle,  and  the  longer  side  of  the  rectangle  ex- 
ceeds the  shorter  side  by  3  in.,  find  the  dimensions  of  the 
rectangle. 

11.  In  a  right  triangle  the  hypotenuse  is  10  ft.  longer  than 
the  longer  leg,  and  the  shorter  leg  is  10  ft.  shorter  than  the 
longer  leg.     Find  all  three  sides. 


QUADRATIC  EQUATIONS  149 

12.  Two  legs  of  a  right  triangle  are  equal.  The  hypotenuse 
exceeds  the  length  of  a  leg  by  3  in.     Find  the  three  sides. 

13.  The  product  of  two  consecutive  integers  is  3540.  Find 
them. 

14.  The  longer  leg  of  a  right  triangle  exceeds  the  shorter 
by  5  in.  The  area  of  the  triangle  is  88  sq.  in.  Find  the 
length  of  each  leg. 

15.  Find  a  number  such  that,  if  20  is  subtracted  from  it, 
and  the  remainder  is  multiplied  by  the  number  itself,  the 
product  is  189. 

16.  A  school  board  has  $  22  to  spend  on  geographies.  If 
the  bookseller  reduces  the  price  of  each  book  5  **,  4  more  books 
can  be  purchased  than  at  the  original  price.  How  many  books 
can  be  bought  at  the  reduced  rate  ? 

17.  Had  a  man's  daily  wage  been  $  1  less,  he  would  have 
been  obliged  to  work  7  days  longer  to  earn  $  140.  How  many 
days  did  the  man  work  ? 

Do  both  roots  apply  to  the  problem  ?     Why  ? 

18.  A  picture  8"  x  12"  is  placed  in  a  frame  of  uniform 
width.  If  the  area  of  the  frame  is  equal  to  that  of  the  pic- 
ture, how  wide  is  the  frame  ?     Draw  a  figure. 

19.  In  a  trapezoid,  AB  exceeds  CD  by  4  in.,  and  CD  exceeds 
the  altitude  CE  by  1  in.  The  area  of  the  trapezoid,  com- 
puted by  taking  the  product  of  the 
altitude  and  one  half  the  sum  of  A 
the  parallel  sides,  is  28  sq.  in.  Find  / 
the  lengths  of  the  parallel  sides  and      /        ! 

of  the  altitude.  A  E  B 

Fig.  29 

20.  The    difference  between  the 

parallel  sides  of  a  trapezoid  is  20  ft. ;  the  altitude  is  equal  to 
the  shorter  of  the  parallel  sides  ;  the  area  is  119  sq.  ft.  Find 
the  lengths  of  the  parallel  sides. 

21.  Find  two  consecutive  integers  whose  product  is  30,800. 


150  ELEMENTARY   ALGEBRA 

22.  Find  two  consecutive  even  numbers  whose  product  is 
1088. 

Hint.  Let  x  be  an  integer ;  then  2  x  must  be  an  even  integer.  How 
much  must  be  added  to  2  x  to  get  the  next  higher  even  integer  ? 

23.  Find  two  consecutive  odd  numbers  whose  product  is 
783. 

Hint.  If  2  x  is  an  even  integer,  what  must  be  added  to  2  x  to  get  the 
next  higher  odd  integer  ? 

24.  The  edges  of  a  rectangular  box  (with  cover)  are  in  the 
ratio  2:3:6.  If  the  entire  surface  is  648  sq.  ft.,  find  the 
lengths  of  the  edges. 

Hint.     Let  the  number  of  feet  in  the  edges  be  2  x,  3  x,  6  x,  respectively. 

SOLUTION  OF  THE  GENERAL  QUADRATIC  EQUATION 

175.  The  equation  ax+bx  +  c  =  0  is  called  the  general 
quadratic  equation,  in  which  the  coefficients  a,  b,  c  may  be 
any  numbers. 

Let  us  solve  it  by  completing  the  square. 

(1)  Divide  both  sides  by  a 

c  .  ,  J    _      c 

and  transpose  _  to  the  right  a,X~~a' 

side. 

(2)  Complete  the  square  by  b        f  b  \2      b2       c 
f  b  V                  •                     x2  -\--x-\- 

adding  (  —  )   to  both  sides. 


\2a) 


a       \2aJ      4  a2     a 


(3)  Perform    the   indicated 
subtraction  on  the  right  side. 

(4)  Find  the  square  root  of 
both  sides. 

(5)  Transpose  — -    to    the 
right  side. 

(6)  Another  way  of  writing  __  —  b  ±  V ft2  —  4  ac 
the  answer.  2  a 


X2-\ X  + 

a 

[2  a) 

62— 4ac 
4  a2 

>    b 
2a 

V&2- 
2 

-  4  ac 
a 

x-       b 
2a 

V&2- 

2 

-  4ac 
a 

QUADRATIC  EQUATIONS  151 

SOLUTION  OF  QUADRATIC  EQUATIONS  BY  FORMULA 
EXERCISES 

176.   1.   Solve  3z2  + 5a; +2=0. 

Here  a  =  3,  b  =  5,  c  =  2.     Substitute  these  numbers  in  the  formula 
I  above.     We  obtain 


x  =  — 

5  ±  V25  -  24  _  -  5  ±  1  _     2Qr       . 
6                         6                3 

Check:  (1) 

3(-§)2  +  5(-f)+2  =  0. 

|--^  +  2  =  0. 

0  =  0. 

(2) 
Solve: 

3(-l)2  +  5(-l)+2=0. 

3-5  +  2  =  0. 

0  =  0. 

2.   5  a;2 -7 

x  +  2 

=  0.                        8.    ±a:2-12a;=10. 

3.  7*2  +  10x-15  =  0.  9    .l*2+z-.3  =  0. 

4.  3  a-2 -a; -10  =  0. 


10.  .3  a:2 +  .2  a- .5  =  0. 

11.  .la?-3z  +  i=0. 
7.   %x2-x-2=0.                        12.   x1  -  101  x  +  7  =  0. 


5.  5a?-6a;-3  =  0. 

6.  z2-10a  +  3  =  0.  "•   -l^-3x  +  i=0. 


MISCELLANEOUS  QUADRATIC  EQUATIONS 
EXERCISES 

177.  Solve  by  the  method  of  factoring,  whenever  the  trino- 
mial can  be  factored  at  sight.  Otherwise  solve  by  the  formula 
or  by  completing  the  square. 

1.  z2- 7  a; -18  =  0  7.   x2  +  24  =  10  x. 

2.  x2-  6  a;  -8  =  0.  x  7 


3.  cc2 +  32=- 18  a\  x  +  60      3x-5 

4.  2a;2  +  5x  +  2  =  0.  9.  a*2 +  3  a;  =  18. 

5.  .2  x2  +  x  —  .5  =  0.  10.  x2  +  7  a?  =  0. 

6.  1.5  a? +  5x  + 1.2  =  0.  11.  40  =  a;2 +  3 a:. 


152  ELEMENTARY   ALGEBRA 

12.    a:2  -  5  x  +  6  =  0.  14.    .1  x2  -  .1  x  =  10. 

13    g  ±  12  g       =26.  15-   *<>  ~  3)  =  0  ~  !)(*  +  *) 

a;  a; +  12       5  16.    a* -8  a; +  15  =  0. 

17.  4x2-7a;-2  =  0. 

18.  (a;-2)(a;-3)  =  (a;-5)(a;  +  l). 

19.  (a;-f)(a;  +  2)  =  0. 

20.  a;(a;  +  i0)  =  (a;-l)(a;-2). 

21.  6a;2 -19a; +  10  =  0. 

22.  a;(a;2  -  x  -  1)  =  (x  -  1)  (x2  +  1). 

23.    7  a?  -  3  x  =  160.  27.   — ?—  +  — ^—  =  0. 

a;2  —  1      a;2  +  1 

28.    -S^  +  i-i. 

x  —  1      a;      3 

=  1.  29.    ^lA  +  ^_  =  9. 

a;  a;  —  1 

30.    -  +  a;  =  5. 
a; 


IMAGINARY  ROOTS 
178.    Solve  a;2  + 2  a; +  5  =  0. 

Complete  the  square,  x2  +  2a;  +  l  =  —  4. 

Extract  the  square  root,  aj  +  l  =  ±V—  4. 

x  =  -1±  V^4. 

We  cannot  express  V—  4  as  a  number  of  the  type  studied 
thus  far  : 

V-4  is  not  +  2,  because    +  2)(+  2)  =  +  4. 

V1^!  is  not  -  2,  because  (-  2)(—  2)  =  +  4. 

V— 4  is  a  new  type  of  number;  it  is  called  an  imaginary 
number. 


24. 

1 

-  10  x  =  3, 

x 

25. 

X 

1 

+  1 

+*4r 

26. 

1 

+  1-0 

QUADRATIC   EQUATIONS  153 

Since  —  4  =(+  4)(—  1),  we  may  write 

-  VT^  VIV^I  =  2V^T. 
The  roots  of  the  quadratic  equation  may  be  written 
x  =  -1  ±2V^I. 

We  callV—  1  the  imaginary  unit.  The  geometrical  mean- 
ing of  imaginary  numbers  will  be  explained  in  the  advanced 
course.  Whenever  a  quadratic  equation  yields  roots  involving 
the  square  root  of  a  negative  number,  it  is  evident  that  some 
condition  is  called  for  which  cannot  be  satisfied  by  the  ordi- 
nary positive  and  negative  numbers. 

For  example,  find  two  numbers  whose  sum  is  2  and  whose 
product  is  3. 

Let  x  =  one  of  the  numbers. 

Then  2  —  x  =  the  other  number, 

and  x(2  —  x)  =  3. 

Solving,  x2  -2z  =  -3, 

x2-2x+  1  =  —  2,_ 
x  -  1  =  ±  V-  2, 
x  =  \±V^2. 
The  solution  involves  V—  2 ;  it  shows  that  no  two  numbers  of  the 
kind  thus  far  studied  can  satisfy  the  conditions.     Only  these  new  num- 
bers, these  imaginary  numbers,  satisfy  the  conditions. 

EXERCISES 
Solve : 

1.  x2  +  x  +  1  =  0.  3.    2  x  -  5  =  x2. 

2.  x2  +  2x  +  6  =  0.  4.    2x2  +  3a;  +  4  =  0. 

GRAPHIC   SOLUTION  OF  QUADRATIC   EQUATIONS 

179.    1.    Solve  graphically  j+2a;-l  =  0. 

x2 
We  draw  the  graph  ofj/=      +  2 x  —  1 .     Assume  different  values  for 
4 

x2 
x  and  compute  2x,  — ,  and  y,  as  shown  in  the  following  table  : 
4 


154 


ELEMENTARY   ALGEBRA 


X 

2x 

X2 

4 

-1 

y 

2 

4 

1 

1 

4 

1 
0 

2 
0 

1 

0 

—  1 

- 1 

-1 

-2 

i 

—  1 

—  2* 

-2 

-4 

1 

—  1 

-4 

-3 

-6 

H 

—  1 

-n 

-4 

-8 

4 

—  1 

-5 

-5 

-  10 

H 

—  1 

-4| 

-6 

-12 

9 

—  1 

-4 

-7 
-8 

-  14 
-16 

12J 
16 

—  1 

-21 
-1 

-9 
-10 

-  18 
-20 

20J 
25 

-1 

n 

4 

y 

A"A 

0 

X 

\ 

\ 

/ 

\ 

Fig.  30 


When  y  =  0,  we  have  —  +  2  x  —  1  =  0.     Hence  values  of  x  in  the  graph 
4 

which  correspond  to  y  =  0  furnish  roots  of  the  equation  —  +  2x  —  1=0. 

We  see  that  y  =  0  at  the  points  J.  and  B.  At  A,  x  =  —  8},  nearly  ;  at  Z?, 
x  =  J,  nearly.  Hence  —  8£  and  +  J  are  approximations  to  the  roots  of 
the  quadratic  equation. 


Solve  graphically: 

2.    ^_2a;  +  l  =  0. 
5 


3.   a?-5a;  +  2  =  0. 


TWO  SIMULTANEOUS  EQUATIONS:    ONE  LINEAR,  THE  OTHER 
QUADRATIC 

180.    I.    Solve  the  two  equations  y  =  x"-, 

y  =  2»  +  3. 

To  find  the  values  of  x  and  y  that  will  satisfy  both  these  equations, 
eliminate  either  x  or  y.  The  elimination  of  y  is  easier.  Writing  in  the 
first  equation  2  x  +  3  in  place  of  y  we  obtain 


QUADRATIC  EQUATIONS 


155 


2x  +  3=x2. 
x2  -  2  x  +  1  =  4. 
x  -  1  =  ±  2. 

a;  =  3  or  —  1. 
From  the  equation  y  =  2  x  +  3,  we  see  that,  when 

x  =  3,       y  =  9, 
and  when  x  =—  1,  y  =  1. 

The  same  values  of  y  are  obtained  with  even  greater  ease  from  the 
equation  „  _  x2 

fx=-l 


The  answers  are 


x~3   and    i 

2/ =  9  U=l. 


It  is  important  to  study  the  geometrical  meaning  of  problems 
like  this. 

In  the  equation 


.'/ 


_  9 


#  +  3, 


give  various  values  to  x  and 
compute  the  corresponding 
values  of  y.  We  obtain, 
say, 

x=2,  1,  0,  -1,  -2,  -3. 
y  =  7,   5,  3,      1,  -1,  -3. 

Plotting  the  pairs  of 
values  of  x  and  y,  as  is 
done  in  Fig.  31,  we  see 
that  all  of  them  lie  on  a 
straight  line. 

In  the  equation  y  =  x\ 
give  different  values  to  x 
and  compute  the  corre- 
sponding values  of  y;  we 
obtain  pairs  of  values,  as 
follows : 

•C    =     Qj  £y  -L« 

y  =  9,     4,     1, 


_.     ^    _      v                                   '     * 

j±    :::;:      '     ±      :±    3J^--- 

:  ±    : :       ::::::      ip      :  (3;9:)i  £  - : 

:  "Tj'iY'flt  ""     *E 

^ "  fiV.Ai                  7 

/:'■<_)                                   a 

Fig.  31 


0,      -1,      -2,      -3. 
0,         1,         4,         9. 


156  ELEMENTARY   ALGEBRA 

These  points  lie  on  the  curve  shown  in  Fig.  31.  This  curve 
is  called  a  parabola. 

We  see  that  the  straight  line  cuts  the  parabola  in  the  two 
points  A  and  B,  having  the  coordinates  x  =  —  1  and  y  =  1, 
x  =  3  and  y  =  9.  These  points  of  intersection  represent  the 
values  of  x  and  y  which  satisfy  both  of  the  two  given  equations. 

II.  Solve  x  =  2y2. 
x  +  2  y  =  4. 

Here  x  is  easier  to  eliminate.  In  the  second  equation  write  2  y2  for  z. 
We  obtain  2  y2  +  2  y  =  4. 

(1)  Divide  both  sides  by  2,  y2+y  =  2. 

(2)  Complete  the  square,  y2  +  y  +  J  =  f. 

(3)  Take  the  square  root  of  both  sides,  y  +  £  =  ±  f. 

(4)  Transposing,  y=±$  —  i  =  l  or  —  2. 

(5)  From  x  +  2  y=4,  we  get  x=  2  when  y  =  1,  and  a;  =  8  when  y  =  —  2. 

The  answers  are  I  *  =  J  and   j  x  =  8 

\y  =  l  lJ/  =  -2. 

Solve  the  following: 

1.  y2  =  4a;,  6.   2a  +  3y  =  0, 
a;  +  2y=5.  x2—Ay  =  0. 

2.  3y2  =  4a>,  7.   y*  +  3x=0, 

2  x  +  4  y  =  14.  6  <e  +  y  =  —  15. 

3.  x2  =  5y,  8.   f+4x  =  l, 

5  y  -  60  =  4  a.  4  a;  -  5  ?/  =  -  23. 

4.  a2  =  10  y,  9.   y2  -  2  x  +  7  =  0, 
2  y  +  7  a  =  90.  x  —  y  =  5. 

5.  3x  =  7y2,  10.   x*  +  3y  +  5  =  0, 
x-y  =  %.  y  +  2x=0. 

PROBLEMS 

181.  1.  The  sum  of  two  numbers  is  18;  the  square  of  one 
number  is  equal  to  3  times  the  other.     Find  the  two  numbers. 

Let  x  =  one  number, 

and  y  =  the  other  number. 


QUADRATIC  EQUATIONS  157 

Then  x  +  y  si  18, 

and  x2  =  3  y. 

Solving,  we  get  from  the  first  equation, 

y  ss  18  —  x. 
Eliminate  y,  x2  =  3(18  —  x)  s|  54  -  3  x. 

Transpose,  x2  +  3  x  =  64. 

Complete  the  square,  x2  +  3  a;  +  (|)2  =  64  +  f  =  *£&. 
Extract  the  square  root,  x  +  f  =  ±  J55-. 

x  =  ±  -V5-  -  f  =  6  or  -  9. 

y  =  18  —  x  =  12  or  27. 

The  answers  are  I x  =  6       and  I  X  =  ~  9 
\y  =  12      dIlu  {y  =  27. 

This  problem  could  have  been  solved  by  the  use  of  only  one  unknown 

quantity. 

Using  both  x  and  y  solve  the  following : 

2.  Find  two  numbers  such  that  10  times  the  square  of  one 
is  equal  to  3  times  the  other,  and  their  sum  is  33. 

3.  Find  two  numbers  whose  difference  is  \,  the  square  of 
the  larger  being  equal  to  f  times  the  smaller. 

4.  Five  times  one  number  is  equal  to  3  times  another.  The 
square  of  the  smaller  is  equal  to  18  times  the  other.  Find  the 
numbers. 

EXERCISES 

182.    1.    Solve  xy  =  6. 
x  —  y  =  5. 

The  second  equation  gives  x  =  5  +  y. 

Substitute  in  the  first,  (5  +  y)y  =  6. 

Or  y2  +  5  y  -  6  =  0. 

Factor,  (y  +  6)  (y  —  1)  =  0. 

y  =-6,  +1. 
Substitute  in  x  =  5  +  y,  x  =  —  1,  when  y  =  —6. 

x  =  6,  when  y  =  1. 

( x  =  —  1  fx  =  6 

The  answers  are  )  and  J 

\irsB-e       U  =  i. 

Let  us  plot  the  equation  x  —  y  =  5  ;  compute  values  as  follows  : 
y  =  2,     1,    0,     -1,     -2,     -6. 
x  =  7,     6,     5,     +4,     +3,     -1. 


158 


ELEMENTARY   ALGEBRA 


The  graph  is  a  straight  line, 
as  shown  in  Fig.  32. 

Plot  next  xy  =  6.    We  have 

y  =  6,      3,      2,      6,      -1, 
-2,     -3,     -6. 

x-l,      2,      3,      1,      -6, 
-3,     -2,     -1. 

The  pairs  of  points  which 
are  here  computed  lie  on 
a  curve  composed  of  two 
branches,  called  hyperbola. 

The  straight  line  and  the 
hyperbola    intersect    in    two 

P^'z^eandj^l, 
FIG-32  z  =  - land  j,  =-6. 

These  pairs  of  values  are  the  answers  to  the  simultaneous  equations. 


\ 

1 

1 

B 

41 

{' 

\ 

\ 

\ 

*S 

£ 

y 

0 

X 

-\ 

7; 

=( 

/f 

P 

/A 

- 

- 

■>> ' 

/ 

Solve  the  following : 

2.  x  +  ?/  =  9, 
xy  =  —  36. 

3.  x  +  y  =  23, 
xy  =  132. 

4.  x  +  y  =  |, 

5.  xy  =  18.75, 
x  —  y  =  11. 


6.  xy+y  =  ±0, 
x-y  =  2. 

7.  xy  -\-  x  =  —  36, 
xy=  —30. 

8.  xy  +  x  +  y  =  0, 

9.  ay  +  x  =  9, 
<c  +  y  =  5. 


10.   Solve  x*+y2  =  25. 
a  +  y  =  7. 

(1)  The  second  equation  gives  x  =  7  —  y. 

(2)  Write  (7  —  y)  in  place  of  x  in  the  first  equation. 

(7  -  j/)2  +  y2  =  25. 
49  —  14  y  +  y2  +  y2  =  25. 

2  y2  —  14  y  =  -  24. 
ya  -  7  y  =  -  12. 


QUADRATIC  EQUATIONS 


159 


The  answers  are 


2/2-7y+(!)2=-12  +  -V  =  }- 

«/=±i  +  l  =  4or3. 
x  =  7  —  j/  =  3  or  4. 
=  4 
3. 


fa;  =  3       J    [x- 


The  graph  of  x  +  y  =  7  can  be  drawn  from  the  following  values : 

x  =  3,     1,    0,     -1,     -2. 

y  =  4,     6,     7,         8,         9. 
These  pairs  of  values  are  points  lying  on  the  line  in  Fig.  33.    • 


V 

S 

, — 

{■■> 

f  } 

/ 

A 

4,' 

5) 

/ 

I 

K 

( 

j 

\ 

j 

>> 

fe 

i> 

/ 

* 

^ 

**\ 

/ 

*o> 

^-- 

y 

-j 

Fig.  33 

To  find  the  graph  of  x2  +  y2  =  25,  assume  values  of  x  and  compute 
corresponding  values  of  y  as  follows  : 

x  =  o,        4,         3,  2,  1,        0,        -1,       -2,     -3,      -4,     -5. 

^=0,     ±3,     ±4,     ±4.6,     ±4.9,    ±5,     ±4.9,     ±4.6,     ±4,      ±3,         0. 

Notice  that  for  every  value  of  x  (except  x  =  5  and  —  5)  there  are  two 
corresponding  values  of  y.    The  pairs  of  points  all  lie  on  a  circle. 


11.    x2  +  y2=  13, 
x-y  =  l. 


12.    x2  +  y*  =  41, 
2x  -y  =3. 


160  ELEMENTARY   ALGEBRA 

13.  x2  +  y2  =  29,  20.    x2  +  yi  =  2, 
x  —  2y  =1.  5x  —  y  =  A. 

14.  3x  +  2y  =37',  21.   ±x  +  y  =  50, 
x>  +  y2  =  113.  x*  +  y2  =  200. 

15.  x*  —  y*  =  0,  22.    7x  +  y  =  17, 
x  +  2y  =15.  a?  +  2/2  =  13. 

16.  a^-2?/2  =  -17,  23.    3  a?  -  ?/2  =  242, 
x  +  y  =4.  a;  —  i/  =  0. 

17.  a;2  +  2?/2=36,  24.    2a^  +  3y2  =  5, 
cc  +  y=6.  x  +  y  =  0. 

18.  2a?-?/2  =  14,  25.   3x2-2yi  =  43, 
x  —  y  =  —  5 .  a;  —  y  =  1. 

19.  a?  +  y*  =  221,  26.    2t/2-4a2  =  82, 
2«-3y=-13.  3x  +  5y  =  Al. 

HISTORY  OF  QUADRATIC  EQUATIONS 

183.  While  the  earliest  known  solution  of  equations  of  the  first  degree 
is  found  in  the  Ahmes  papyrus,  about  2000  b.c,  the  earliest  known  solu- 
tions of  quadratic  equations  occur  in  Euclid's  Elements  of  geometry, 
about  300  b.c,  —  about  1700  years  later.  Euclid  solved  quadratic  equa- 
tions by  geometric  construction.  In  proposition  11,  of  Book  II,  of  the 
Elements,  Euclid  solved  by  drawing  lines  the  problem  :  To  divide  a 
given  straight  line  into  two  parts,  so  that  the  rectangle  contained  by  the 
whole  line  and  one  part  of  it  may  be  equal  to  the  square  on  the  other  part. 
In  algebra,  this  problem  demands  the  solution  of  the  quadratic  equation 
a(a  —  x)  —  a;2,  where  x  is  the  length  of  the  required  line.  About  two 
centuries  after  Euclid,  a  Greek  writer,  Heron,  gives  a  root  of  a  quadratic 
equation,  showing  that  he  knew  how  to  solve  the  equation  algebraically, 
but  he  nowhere  tells  how  he  did  it.  Equally  non-committal  is  Diophan- 
tus,  the  celebrated  Greek  writer  on  arithmetic  and  algebra,  of  the  fourth 
century  a.d.  Diophantus  gives  correct  answers  to  several  quadratic 
equations.  Part  of  his  book  is  lost ;  the  parts  that  are  extant  nowhere 
explain  his  mode  of  solution.  Diophantus  does  not  give  more  than  one 
root  of  a  quadratic  equation,  even  if  both  are  positive,  nor  does  he  ever 
recognize  a  negative  root.  He  nowhere  uses  negative  numbers.  In 
that  respect  his  algebra  was  like  our  school  arithmetics. 


QUADRATIC  EQUATIONS  161 

Marked  advance  in  algebra  was  made  by  the  Hindus.  In  the  fifth 
and  sixth  centuries  they  solved  quadratics  by  a  process  which  can  be  ex- 
plained easiest  by  using  our  modern  notation.     They  wrote  the  roots  of 

ax2  +  bx  =  c;  thus :  x  = — - Somewhat  later  Cridhai-a,  an- 

■\/4  ac  +  6^  —  6 

other  Hindu  mathematician,  uses  the  form  of  x  = •  Toob- 

2a 

tain  the  first  form  of  x  one  begins  by  multiplying  both  sides  of  ax2  +  bx  =  c 
by  a.  To  obtain  the  second  form  of  x,  one  begins  by  multiplying  both 
sides  of  the  equation  by  4  a.  This  last  process  is  now  called  the  "  Hindu 
method"  of  solution.  It  has  the  advantage  of  excluding  fractions  under 
the  radical  sign. 

The  Hindu  mathematician  Bhaskara  (twelfth  century  a.d.)  recognized 
that  a  quadratic  equation  has  two  roots,  when  both  of  them  were  positive. 
He  gives  x  =  50  or  —  6  as  the  roots  of  x2  —  45  x  =  250,  and  remarks  : 
"  But  the  second  value  is  in  this  case  not  to  be  taken,  for  it  is  inadequate  ; 
people  do  not  approve  of  negative  roots."  Thus  negative  roots  were  seen, 
but  not  admitted.  The  recognition  of  negative  roots  came  much  later. 
The  Italian  mathematician,  Cardan  (1539),  and  the  Belgian  mathematician, 
Simon  Stevin  (1585),  admitted  negative  roots.  Soon  after,  Albert  Girard 
( 1629)  took  the  still  more  advanced  position  of  admitting  the  possibility  of 
imaginary  roots,  thereby  recognizing  the  theorem  that  every  quadratic 
equation  has  two  roots  or,  more  generally,  that  every  equation  of  the  nth 
degree  has  n  roots.  But  opposition  to  the  acceptance  of  imaginary  roots 
continued  for  over  a  century  after  Girard.  Not  until  the  beginning  of 
the  nineteenth  century  did  all  opposition  to  them  vanish.  This  opposition 
to  imaginary  roots  grew  mainly  out  of  the  fact  that  they  could  not  be 
explained  geometrically .  A  positive  root  could  be  pictured  to  the  eye  by 
the  length  of  a  line  drawn  in  a  given  direction  ;  a  negative  root  could  be 
interpreted  by  the  length  of  a  line  drawn  in  the  opposite  direction.  But 
how  could  imaginary  roots  be  consistently  represented  by  lines  ?  Such 
geometric  representation  is,  however,  possible.  How  this  can  be  done 
was  first  shown  by  the  Danish  engineer,  C.  Wessel,  in  1797.  Most  influ- 
ential in  securing  the  general  recognition  of  imaginary  numbers  was  the 
great  German  mathematician,  Gauss. 


CHAPTER   XIV 

FRACTIONS 

184.  We  have  seen  in  previous  exercises  that  fractions  in 
algebra  are  subject  to  the  same  general  rules  as  fractions 
in  arithmetic.  In  algebra,  as  in  arithmetic,  the  value  of  a 
fraction  is  not  changed,  if  both  of  its  terms  are  multiplied  by  the 
same  factor  or  if  both  of  its  terms  are  divided  by  the  same  factor. 

2     4     a     ac     ax  +  bx     a  +  b 


Thus, 


3     6     b     be     ax  —  bx     a  —  b 


~2  _  5 
The  most  common  mistake  made  by  beginners  is  to  say  that  is 

equal  to  — .   But  this  answer  cannot  be  obtained  by  multiplying  or 
6 

dividing  both  terms  of  the  given  fraction  by  any  number. 

EXERCISES 

185.    1.    Is  ®^zA  equal  to  —  when  x  =  1  ?   When  x  =  2? 
a2 +  6    H  6 

2.  Show  by  an  example  that  subtracting  a  number  from  the 
numerator  and  denominator  changes  the  value  of  a  fraction. 

3.  Which  of  the  following  equalities  are  correct  ? 

3  +  l_,    1  (a  +  b)(a  -b)  __  a  +  b       a  +  a? ^a 

3-2      -2'    (o  +  2  6)(a-6)      a +  26'    b  -  a2     b 

4.  Reduce  to  the  lowest  terms,  ^-^ ^~- 

'  16  ary- 32  xy 

Factor  numerator  and  denominator,  then  divide  both  terms  by  their 
common  factors.     We  get 

3 
12  a8ys(s2  -  4  y2)  _  n#Pg*(x  +  2y)  (&-2jfr  _3(x  +  2y) 
16a^y5(x  —  2  y)  IS-x^i^z-* — S-57  4x 

4x 
162 


FRACTIONS  163 

Reduce  to  the  lowest  terms  : 


*\         1.5.6.  19 


abx 


6(g  -  6)(6  -  c) 
(3  6-3a)(4  6-4c) 


6.  ^-  20     a2-2a~3 
™>V  a -3 

7.  «^.  2       a2+a6-662 
«**y  '     5(a-2  6)  ' 

&    a(x  +  y)  §  22         2a2&(a3-&3) 

a(»  +  y)  '    6  a6(a2  +  ab  +  62) 

o     ac2  +  yc2  23  m2-l 


^2  -  2/c  '  (1  -  m)(w»  +  4m  +  3)' 

10.    *(*  +  !)' 24         (2y-l)2 

2a?(aj-l)(a?  +  l)  *  '  2f-  5y  +  2 

11.  3a:  +  3y  25  n3  -  n2  -  12 n  | 
4a  +  4i/  '       4rc2-16n 

12     «26  +  q62  26  ax -ay 


(y  -  xy 


13.  *  +  *.  27     5a~10&- 

y  +  «y  2  b  -  a 

14.  fr-fP.  28     w^+my^ 

2/  —  *  nx  +  ny 

15.  *=*.  29.         5a  +  56 


J/  —  a?  15  a2&  +  15  o& 

i6.    ^-y8  30    y2-5y  +  4 
(*  +  y)2  y3  -  y 

17.  12x*y(x  +  2)(x  +  4:)_  M     4^-4^  +  y' 

9  #?/2(a;  +  4)2  ay  —  2  ax 

18.  (y-«)(ya  +  «2)  32.    (3y-aQ», 

x2—  y2  2x  —  6y 


164  ELEMENTARY   ALGEBRA 

MULTIPLICATION   AND  DIVISION  OF  FRACTIONS 

186.  In  algebra  we  have  the  same  rules  of  operation  as  in 
arithmetic  ;  namely, 

1.  In  finding  the  product,  multiply  the  numerators  together  for 
a  new  numerator,  and  the  denominators  together  for  a  new  de- 
nominator. 

2.  In  finding  the  quotient,  invert  the  divisor  and  then  proceed 
as  in  the  multiplication  of  a  fraction  by  a  fraction. 

Thus,  |  +  ?  =  §  x  |  =  }f 

The  correctness  of  these  rules  is  shown  in  arithmetic.  How- 
ever, it  is  easy  to  establish  the  truth  of  them  by  means  of  the 
equation. 

For  instance,  the  second  rule  may  be  proved  as  follows  : 

Let  ^  =  x,         ?  =  y, 

b  d 

then  a  =  bx,        c  =  dy, 

and  ad  =  bdx,     be  =  bdy. 

Dividing  equals  by  equals, 

ad  _  bdx  _  x 

be      bdy     y 

x     a      c 

y~b^d 

a     c      ad 

b  '  d~~  be 

This  proves  the  rule  for  the  division  of  one  fraction  by 
another. 

Notice  that  the  rules  for  the  multiplication  and  division  of 
fractions  are  still  applicable  when  in  place  of  one  of  the  frac- 
tions we  have  an  integral  expression  ;  we  need  only  write  the 
integral  expression  with  the  denominator  1. 

Thus,  «.c=?.^  =  ^-c. 

6  6     16 

6  6      1      6     c     6c 


But 
Hence 


FRACTIONS  165 

Perform  the  indicated  operations  and  simplify  by  reducing 
the  resulting  fraction  to  its  lowest  terms : 

1. 

3       31 

iaer.^5      31       31 
Here  ==-t==t-^=S 


120 
YT5 

.155 
T5TP 

3       31 

oaer.j^s      31 
2&s-ym    4-17 

£1       4 
17 

X2- 

-y-   . 

x  +  y 

x2  —  4  y2      x—  2y 

Inverting  the  divisor  and  multiplying,  we  obtain 

(x2-y2)(x-2y) 
(x-2  _  4y2)(x  +  y) 

It  is  usually  best  not  to  carry  out  the  indicated  multiplications  in  the 
numerator  and  denominator.  On  the  contrary,  it  is  usually  better,  if 
possible,  to  factor  still  further,  with  the  purpose  of  exposing  to  view  all 
the  factors  that  are  common  to  the  numerator  and  denominator. 

Factoring  and  then  dividing  both  terms  by  their  common  factors,  we 

(s-f-yMar  -  y)Cz-=-3ir)   _x-y 
(a;  +  2y)(j^--3-y)Cz-Hr)      x  +  2y 

„       a  —  b  _^  a2  —  ab  1  ft     15  m  —  30         3  m 


4.   ^L-f.^L.  11. 


12. 


6.    -^-.=-=^.  13. 


2  m  5  m  -  10 

y2  —  x- .  y2  +  x2 

x      xy  —  x2 

7  x  —  14y  ___  7(x— 2  y) 

5~~  4 

x3  +  J/3  ^_  x  +  y 
xt/  x2 

7     *~  y2       6s  x3  — y3  .  x2  +  xy  +  ,y2 

'5  20 

O 

n    a2  —  1      a  4- 1  ,  -    /  2  .     2\     x2w  +  v3 

9.    ___- 1._^__.  16.   (x2  +  y2)-n    •  3  ^  • 


166  ELEMENTARY   ALGEBRA 


"■  r¥h+Sr-  19-  v-v+n-^- 

l  +  8r     l+2x  x*  +  y3 

18.      6a?      •  1-3^.  20.    3x  +  1  •  (a>  +  y). 


21. 

22. 
23. 
24. 


2s2  +  5a:  +  3  _  2a?2  +  s-3 
x  —  1  x  +  1 

2 a2  +  a;  —  6  a  -\-x 

ax  +  x2      '  2  x2  +  7  a;  -  15 ' 

4y2  +  6y  +  2.      2a;,y  +  2a; 
3  afy  +  6  a?    '  6  y2  + 15  y  +  6  * 

x2  —y2        a?  + 1        a?  —  1 
a;  + 1      a?  +  a:  + 1     a;  —  y 

COMPLEX  FRACTIONS 

187.  A  complex  fraction  can  usually  be  simplified  most 
easily  by  multiplying  both  its  numerator  and  denominator  by 
the  lowest  common  denominator  of  the  fractions  occurring  in 
the  numerator  and  denominator  of  the  complex  fraction. 

a     b 

1.    Simplify    . si- 

x     y 

The  1.  c.  d.  of  the  minor  fractions  -  and  -  is  xy.     Multiply  both  the 

x         y 

numerator  and  the  denominator  of  the  complex  fraction  by  xy.     We 

obtain  . 

a  ,  o 

x     y  _ay  +  bx 

a  _b~  ay  —  bx 

x     y 


2.    Simplify 


4     5_6 
a     b     c 
1_3_4' 
a     b     c 


FRACTIONS  167 


Multiply  both  terms  of  the  complex  fraction  by  abc. 
Then, 


4      5_6 

a     b     c      4  be  +  5  ac  —  6  db 


3.    Simplify 


1_3_4       6c  -  3  ac  -  4  ao 
abc 

25 

49 

35' 

42 

In  this  case  the  process  explained  in  arithmetic  is  easier  | 
Invert  the  divisor  and  multiply. 

25 
49 
35 
42 
Simplify: 

2 1_ 

t+t 

5x     1  y 

m  _ZL 
5.     "      m.  8.  2m  +  1 

»     m  - 


5 

7 

6 

-35 

7 

30 
49 

3 
4m 

3n 

4  + 

1 

mn 

10. 


1  + 


m      n  2m+l 

wi  «      5      6 

6-  JIT  sir 

m  60     20                                   <c  +  2 


22 
63 

55* 

72 

1+- 

x-1- 

2 

x  +  2 

x-2- 

3 

1    ■+:    1 


13. 


1—x      1+y 

JL 1_' 

1  — a?     1  +  y 


168  ELEMENTARY   ALGEBRA 

THE  HIGHEST  COMMON   FACTOR 

188.  Only  expressions  which  are  free  of  fractions  and  of 
radicals  will  be  considered  here. 

The  highest  common  factor  (h.  c.  f.)  of  two  or  more  expres- 
sions is  the  product  of  such  prime  factors  as  occur  in  each  ex- 
pression, every  factor  being  taken  the  least  number  of  times  it 
occurs  in  any  one  expression. 

This  definition  indicates  the  process  of  finding  the  h.  c.  f. 

1.  Find  the  h.  c.  f.  of  12  a2b3c,  24  a36c2,  36  aW. 
Resolving  into  prime  factors,  we  obtain 

12  a2b*c  =  2-2-3-a-a-b-b-b-c. 
24  u86c2  =2-2-2S-a-a-a-b-c-c. 
36  c462c8  =  2-2-3. 3-  aa-a-a-b-b-c-c-c. 
The  h.  c.  f .  =  2    2  •  3  •  a  •  a  •  b  ■  c  =  12  a?bc. 

2.  Find  the  h.  c.  f .  of  a\a  -  2)2,  ab(a  -  2)3,  3  a{a  -  2)4. 

We  may  write 

a2(a-2)2  =a-a(a-2)(a-2). 
ab(a  -  2)3  =  a  •  b(a  -  2)(a  -  2)(a  -  2). 
3 a(a  -  2)*  =  3  -  a(a  -  2)(a  -  2)(a  -  2)(a  -  2). 
.-.  theh.c.  f.  =  a(a-2)(a-2)  =  a(a-2)2. 

EXERCISES 

189.  Find  the  h.  c.  f.  of: 

1.  a2b,  a%  a2d.  8.  2  a  +  2  b,  a2  -  b2. 

2.  axy,  bxy,  cxy.  9.  x  —  y,  x2  —  y2. 

3.  10  a2,  15  a3,  25  a4.  10.  x2  -  1,  x2  -  2  a  +  1. 

4.  6  a26c,  12  aW,  18  aW.  11.  2  a2  -f  4  a  +  2,  4  a2  -  4. 

5.  a2  —  a&,  a2  —  ac,  a2  +  ad  12.  3  ab  +  3  &2,  a2  +  2  a&  +  b\ 

6.  mx—my,  nx—ny,  px—py.  13.  cc2  —  2  a;?/  +  y2,  x3  —  y3. 

7.  2  z2  +  2#2,  ax2  +  ay2.  14.  a2  -  4,  a2  +  a;  -  6. 

15.  x3  +  y3,  x2  —  y2,  x2  +  2  xy  -\-  y2. 

16.  m2  —  9,  m2  +  5  m  +  6,  m2  —  m  —  12. 

17.  a2  -  1,  a3  -  1,  a&  -  &  -  2  a  +  2. 


FRACTIONS  169 

18.  3  a6  -  48  a2,  3  a4  +  13  a3  +  14  a2. 

19.  (*  +  2/)4(x  -  y)\a  +  b),  (x  +  y)\x  -  y)\a  -  b). 

20.  r2  -  6  rs  +  8  s2,  r2  +  2  rs  —  8  s2,  r2  —  5  rs  +  6  s2. 

LOWEST  COMMON   MULTIPLE 

190.  The  lowest  common  multiple  (1.  c.  m.)  of  two  or  more 
expressions  is  the  product  of  all  the  prime  factors  that  occur 
in  the  expressions,  e%rery  factor  being  taken  the  greatest  number 
of  times  it  occurs  in  any  one  expression. 

This  definition  indicates  the  process  of  finding  the  1.  c.  m. 

1.  Find  the  1.  c.  m.  of  3  a2,  6  a3,  5  ab2. 

3a2  =  3-a-a;  6  as  =  2 -3  ■  a  ■  a  •  a;  5a62  =  5-a-6-6 
Hence  the  1.  c.  m.  =  2  •  3  •  5  •  a  ■  a  ■  a  .  b  ■  b  =  30  a862. 

2.  Find  the  1.  c.  m.  of  a2  +  2  ab  +  b2,  a2  -b2,  2  a2  +  2  ab. 

a2  +  2ab  +  6*  =(a  +  b)(a  +  b). 
a2-  b2  =  (a  +  6)(a-6). 
2a2  +  2a6  =  2a(a  +  6). 
Hence  the  1.  c.  m.  =  2a(a  +  b)2(a  —  b). 

EXERCISES 

191.  Find  the  I.e.  m.  of: 

1.  x,  y,  z.  5.  a2,  a2  -|-  a. 

2.  a?b,  ab3.  6.  a2  +  a,  a2  —  a. 

3.  b2cd,  bc2d,  bed2.  7.  c2  +  cd,  cd  +  d2. 

4.  5  c,  7  d,  10  e.  8.  a2  4-  a&,  a&  —  b2. 

9.  a2  —  4,  a2  -f  a  —  6. 

10.  x2  —  y2,x2  +  2xy  +  y2,  (x  -  y)9. 

11.  a2  -  b2,  a3  -  b3,  a2  -  2  ab  +  b2. 

12.  m  +  n,  m2  +  n2,  m2  +  2  mn  +  «2- 

13.  3  m  —  3  n,  m3  —  n3,  m2  +  raw  +  n*. 

14.  s2  +  6  x  +  9,  a2  +  5  x  +  6. 


170  ELEMENTARY   ALGEBRA 

15.  a2-6a  +  8,  a2 -a -12. 

16.  1  —  x2,  1  —  x3,  1  +  x. 

17.  xyz,  x2y  —  xy2,  x*y  —  xy3. 

18.  x2  —  9,  a  +  3,  x2  +  10  a  +  21. 

19.  a2  -  1,  a3  +'a2  +  a  +  1,  a3  -  a2  +  a  -  1. 

20.  x2  —  £  xy  +  4:  y2,  x2  +  2  xy  —  8  y2,  x2  —  xy  —  2  y2. 

192.  Find  the  h.  c.  f .  and  1.  c.  m.  of : 

1.  24,  36.  6.  25,  75  cd4,  100  ac*d. 

2.  96,  144.  7.  3  a,  a2  -  ab. 

3.  49,  98,  147.  8.  4  aa,  4  a3^  -  4  aa?. 

4.  252,216,180.  9.  x2-9,x2+5x  +  6. 

5.  8  a*/3,  4  artyz,  10  tfyh?.  10.  as  +  ab,  bx  +  b\ 

11.  x2  —  y2,  x2  +  2  xy  +  y2,  x*  -{-  y3. 

12.  27  -  a3,  (3  -  a)2,  4  (3  -  a). 

13.  a2  -  2  a&,  a2  -  6  aft  +  8  b2,  a2  +  ab-6  b2. 

14.  c2  —  cd  —  2  d2,  cm  +  dm  —  en  —  dw. 

ADDITION  AND  SUBTRACTION  OF  FRACTIONS 

193.  The  process  is  the  same  as  in  arithmetic.  If  the 
fractions  do  not  all  have  the  same  denominator,  they  must  be 
reduced  to  fractions  having  a  common  denominator.  The 
lowest  common  denominator  (1.  c.  d.)  is  obtained  by  finding  the 
lowest  common  multiple  (1.  c.  m.)  of  the  given  denominators. 

1.    Add  — —  +-^— . 

2a&3       3a26       4a262 

The  1.  c.  d.  of  2  a68,  3  a26,  4  a262  is  12  a268. 

12 a268  ♦  2 a63  =  6 a,  6a-    6oa! 


12  a268 -*- 3  a26  =  4  6s,    _ 
12a263H-4a262  =  36, 


X 

2  aft3 

6a 
'  6a 

X2 

3a26* 

4  62 
4  62 

X3 

36 

4a262 

36 

12  a263 
4  62x2 
12a263' 
3  6x3 
12a268" 


FRACTIONS  171 

We  obtain^ xj_      j?g_  =  6  «x  -  4&2x2  +  3^ 

2aba     3a2b      4a2b2  12  aW 

In  practice,  much  of  this  work  can  be  done  mentally  and  need  not  be 
written  down. 

2>  2x         | 6y 

(x  -  y)\x  +  y)      (x  -y)(x  +  y)2'     , 
The  1.  c.  d.  =  (x  -  y)2(x  +  y)2. 

(x  -  y)2(x  +  y)2  divided  by  (x  -  y)2(x  +  y)=x  +  y. 
(x  —  y)2{x  +  y)2  divided  by  (x  —  y)(x  +  y)2  =  x  —  y. 

Hence  2x  +  6y  =  2s(a  +  y)  + 6y(s  -  y) 

{x-y)\x  +  y)      (x-y)(x  +  y)2  (x-y)2(x  +  y)2 

_  2  x2  +  8  xy  —  6  y2 
(x-y)2(x  +  y)2' 

194.  EXERCISES 

1.  Add  #,  I,  JL,  3.   Add  — ,  -^-,   — — . 

*  *'  **  a;2?/'  3xy2'  15afy2 

2.  Add  »    JL,  Z*.  4.   Add  «5,  _«*   l*¥. 

5    15'  25  5  '       25'     50 

Change   to  a  single  fraction  and  reduce  that  to  its  lowest 
terms : 

5.  *±2_£zJ!.  12.        1     +_!_. 

5  10  x+y      x—y 

6.  J L-.  13.         »  » 


x+y     x—y  x  —  y     x+y  > 

7.         2      '+-A-.  .    14.  m       +  ■     * 


(x  —  y)2      x—y  x(m  —  x)      m  —  x 

8     _a l  15      x  —  y  x 

x  +  y      {x  +  yf  '    x2  +  xy     x2  —  y2' 

9.     _* L_.  16.    -       4        + 


X2  _  y2     J,  _j_  y*  e  x  _|_  15     8  x  +  20 

io.      y    +-1-.  it.  -£^ y~2  . 

x(x  +  y)      x  +  y  15  y  —  5     18  y  —  6 

11     _x^-jj_  _^_x_+j_  lg       s  +  2         a;  -  2 
2x  +  2y  '  3a;-3#"  '    5z  +  5      lx  +  l' 


172  ELEMENTARY   ALGEBRA 


19.         30*     + 


9a?-l     3s-l     3z  +  l 

20.  — 1 + ! + 1 . 

(s  -  «)(*  -y)     (*  -  y)0  -  *)     (y  —  «)(z  -  a?) 


(a  +  6)  (a  +  c)      (a  +  6)(6  +  c)      (6  +  c)(a  +  c) 

EQUATIONS  CONTAINING  FRACTIONS 

195.  In  solving  equations  containing  fractions  it  is  usually 
best  to  multiply  both  sides  of  the  equation  by  an  expression 
which  removes  all  fractions. 

1.  Solve -  +  ?—?  =  -. 

6         3         3 

Multiplying  both  sides  by  6,  x  +  2  x  —  4  =  14. 

3x  =  18. 
x  =  6. 
Check :   Substituting  6  for  x  in  the  original  equation, 
I  +  1-* 

2.  Solve  1+     ' 


x      x  -4- 1      x—1 

The  Led.  =  x(x  +  l)(x  —  1).     It  is  better  to  retain  the  Led.    in 
the  factored  form. 

Multiplying  both  sides  by  x(x  +  l)(x  —  1),  we  obtain 
(x  +  l)(x-l)+x(x-  l)  =  2x(x  +  1). 
x2  -  1  +  x2  -  x  =  2  x2  +  2  x. 
-3x  =  l. 
x=-J. 

Check:  -3  +  \  =  -^-. 

I      - 1 

-3  +  *=-*.. 


DOT  3,4  11 

3.    Solve h 


x2  -\-x     x  +  1      6x  —  6 
The  Led.  =  6x(x+l)(x-  1). 


FRACTIONS  173 

Multiply  both  sides  by  the  1.  c.  d. 

18  x  -  18  +  24  x2  -  24  x  =  11  x2  +  11  x. 
13  x2  -  17  x  -  18  =  0. 
(x-2)(13x  +  9)  =  0. 

x  =  2  or  -  ft. 

EXERCISES 
196.    Solve  the  following : 

1  -  =  I.  14      g-1  =  3x-4 

"    x     8*  '    2x-3     6x-5* 

2  2  =  J5_#  15.    |x-.5x  =  f. 
3      2x  2x*  x 

lb. 


6. 


7'      x       12 


11. 


12. 


4  6 

3x-l     1 


x  +  7       5 
13.    2x  +  4,(4  +  x)=3§. 


3     2x-l      x-2_2  2x2+l      a;  - 1 

2  3     ~3*  17     _2 L*»S, 

x_-fl      x=13  *-!      2-x     x 

3  4       4'  18.         1       |       2      =g 

5.     §x  +  5x  =  7.  x-1      x-2     6 

6-x         4     _4  19.       2     +     3     =Q. 


3-x     3-x     3 

0.6  _  7  20.    y  +  2  = 


x  —  1      x  — 2 
25 


y  +  2" 


3x  1  21.    x—^  +  -^-  =  0. 


x  +  2     x— 2 
13  x        14  x  22. 


X4-1        X— 1 

x+1         3 


9.     _±^L  =  J^L.  x-1       X-3 
x  +  4      x-f-5 

00  x  — 5         4         3x—  1 

.001.**  23.  __  =  _+___. 

x  +  2       24 


75  24-  fri  =  2a;+1- 


25. 

x+12  =  _36. 

X 

26. 

1         _x# 
x—  4         4 

174  ELEMENTARY   ALGEBRA 


27.    -JL+     2-te§|. 


a?  - 1     a?  —  2     90  z  +  1     2  +  2     z+3 

28.    ^±i+izL^=§.       30.    ^-^+l+^+^+l=2a.+3, 
a  — 1     3  +  x     x  x—1  x+1 


PROBLEMS 

197.  1.  By  what  number  must  72  be  divided,  that  the 
quotient  may  exceed  the  divisor  by  6  ? 

2.  What  number,  when  divided  by  13,  yields  a  quotient 
that  is  less  than  the  number  by  168  ? 

3.  A  is  8  years  old,  B  is  33  years  old.     When  will  A  be  ^ 
as  old  as  B  ? 

4.  A  is  36  years  old,  B  is  48  years  old.     How  many  years 
ago  was  A  i  as  old  as  B  ? 

5.  What  fraction,  equal  to  ^,  becomes  equal  to  f  when  15  is 
added  to  its  numerator  and  denominator  ? 

Hint.    Let  —  be  the  fraction. 
Sx 

6.  What  must  be  the  dimensions  of  a  rectangle  containing 
192  sq.  in.,  in  order  that  the  perimeter  be  56  in.  ? 

7.  A  rectangle  whose  area  is  ^  sq.  ft.  has  a  perimeter  of  If 
ft.     Find  its  dimensions. 

8.  Divide  84  into  two  parts,  such  that  the  fraction  formed 
by  these  parts  is  f . 

9.  A  number  exceeds  the  sum  of  its  one  third,  one  fourth, 
and  one  fifth  by  26.     Find  the  number. 

10.  The  rate  of  one  train  running  between  two  towns  exceeds 
the  rate  of  another  by  10  miles  an  hour.  The  difference  in 
time  is  1  hour.  Find  the  rate  of  each  train,  if  the  towns  are 
200  miles  apart. 

Negative  answers  for  the  rates  may  be  rejected  as  not  permissible  in 
our  problem. 


FRACTIONS  175 

11.  An  oil  tank  can  be  filled  by  one  pump  in  6  hours,  or  by 
another  pump  in  12  hours.  How  long  will  it  take  to  fill  the 
tank  when  both  pumps  are  working  ? 

Let  x  =  time  in  hours,  when  both  pumps  are  working. 

Then         -  =  part  of  tank  filled  by  both  pumps  in  1  hr. 

I  =  part  of  tank  filled  by  the  first  pump  in  1  hr. 
■fa  =  part  of  tank  filled  by  the  second  pump  in  1  hr. 

Hence  1+1  =  1.    Solve. 

6      12      x 

12.  How  long  will  it  take  two  pumps  to  fill  an  oil  tank, 
when  one  pump  alone  can  fill  it  in  6  hr.,  the  other  pump  alone 
can  fill  it  in  8  hr.  ? 

13.  Two  pumps  working  at  the  same  time  can  fill  an  oil 
tank  in  5  hr.  One  pump  working  alone  can  fill  it  in  8  hr. 
How  long  will  it  take  the  second  pump  alone  to  fill  it  ? 

14.  In  a  debating  society  a  motion  was  carried  4  to  3 ;  on  a 
reconsideration,  4  affirmative  votes  changed  over  to  the  nega- 
tive, and  the  motion  was  lost  10  to  11.  Find  the  number  that 
voted  in  the  affirmative  on  the  first  ballot. 

Hint.  Let  the  affirmative  and  negative  vote  on  the  first  ballot  be  4.x 
and  Sx,  respectively. 

15.  Find  the  world's  production  of  rubber  in  a  year,  when 
South  America  produced  |,  Africa  \  of  the  total  amount,  and 
the  rest  of  the  world  produced  2800  tons. 

16.  A  man  spends  \  of  his  income  in  board,  TL  in  clothes,  £ 
in  sundries,  and  has  $  630  left.     What  is  his  income  ? 

17.  A  man  placed  at  interest  $6000  at  5%,  and  6  months 
later  $5500  at  6  %.  When  will  the  interest  on  the  two  sums 
be  equal  ? 

18.  Two  ball  teams  in  a  league  have  the  following  record: 
A  won  50,  lost  24,  B  won  56,  lost  24  games.  The  two  teams 
play  a  final  series  of  10  games  together.  How  many  must  A 
win  in  order  that  the  ratio  of  games  won  to  games  lost  be 
equal  for  both  teams  ? 


176  ELEMENTARY  ALGEBRA 

19.  The  height  of  a  bin  is  1  foot  less  than  the  width  and  4 
feet  less  than  the  length.  Find  the  capacity  of  the  bin  if  the 
height  is  equal  to  half  the  length. 

HISTORY  OF  FRACTIONS 

198.  Fractions  occur  in  some  of  the  earliest  historic  records,  but 
ancient  peoples  experienced  great  difficulty  in  computing  with  fractions. 
At  the  present  time  fractions  may  have  any  number  as  numerator  and 
any  number  as  denominator.  Not  so  in  olden  times.  The  Babylonians, 
as  early  as  2000  b.c,  perhaps  even  earlier,  used  fractions  with  the 
denominator  60.  In  measuring  time  and  angles,  they  used  subdivisions 
on  the  scale  of  60.  The  subdivisions  of  the  hour  into  60  minutes  and  the 
minute  into  60  seconds,  as  also  of  a  degree  (angular  measure)  into  60 
minutes  and  a  minute  into  60  seconds,  are  of  Babylonian  origin.  Every 
time  we  measure  angles  and  every  time  we  consult  our  watches  we  are 
using  units  of  measure  invented  by  astronomers  near  the  banks  of  the 
Ganges  over  3000  years  ago  !  When  the  Babylonians  took  60  as  the  de- 
nominator of  all  fractions,  there  was  no  need  of  actually  writing  the 
denominator.  They  wrote  simply  the  numerator.  To  distinguish  it  from 
a  whole  number,  they  placed  it  a  little  to  the  right  of  the  ordinary  position 
for  a  word  or  number. 

A  fractional  system  involving  the  same  denominator  was  in  vogue  also 
among  the  Romans.  They  usually  took  12  as  the  denominator;  with 
them  12  was  also  the  number  of  subdivisions  of  weights  and  measures ; 
the  coin  called  as  was  subdivided  into  12  uncice,  the  foot  into  12  inches. 

While  the  Babylonians  and  Romans  preferred  the  use  of  fractions  with 
the  same  denominator  and  different  numerators,  the  Egyptians  and 
Greeks  had,  as  a  rule,  fractions  with  constant  numerators  and  different 
denominators.  In  the  Ahmes  papyrus,  an  old  Egyptian  mathematical 
treatise  found  about  fifty  years  ago,  much  attention  is  given  fractions. 
All  fractions  in  that  papyrus,  except  the  fraction  f ,  have  the  numerator 
unity.  In  this  case  there  was  no  need  of  writing  the  numerators ;  a 
fraction  was  designated  by  writing  the  denominator  and  then  placing 
over  it  a  dot  or  another  simple  mark.  Two  ninths  was  indicated  in  the 
Ahmes  papyrus  as  £  fy,  the  sum  of  £  and  ^  being  }. 

The  more  modern  point  of  view  in  computation  with  fractions  appeared 
among  the  Hindus  after  the  fifth  century  of  our  era  and,  somewhat  later, 
among  certain  Arabic  scholars,  and  among  European  writers  of  the 
renaissance.    The   Hindus  wrote  the  numerator  over  the  denominator, 


FRACTIONS  177 

but  did  not  separate  the  two  by  a  fractional  line.  Among  the  first  to  use 
this  line  was  the  Italian  mathematician  Leonard  of  Pisa  (thirteenth 
century).  At  the  beginning  of  the  sixteenth  century  the  fractional  line 
had  come  into  general  use. 

Even  as  late  as  the  fifteenth  century  certain  European  writers  of 
prominence  experienced  difficulty  in  explaining  the  multiplication  of 
fractions.  If  to  "  multiply  "  means  to  "increase,"  how  can  the  product 
of  two  proper  fractions  be  smaller  than  either  of  the  two  fractions  ?  The 
earliest  algebraist  to  make  extensive  use  of  letters  as  the  representatives 
of  numbers  was  the  Frenchman,  Vieta  (1540-1603).  The  use  of  fractions 
involving  letters  was  introduced  since  his  day. 

The  most  conspicuous  figure  in  the  invention  of  decimal  fractions  is  the 
Belgian  mathematician,  Simon  Stevin.  He  explained  them  in  a  booklet, 
La  Disme,  published  in  1585.     He  did  not  use  the  decimal  point.     He 

0  12     3 

had  a  clumsy  notation ;  he  wrote  the  fraction  5.912  thus  5  9  1  2  or  thus 
5®9®1®2@.  The  small  digit  above,  or  inside  a  circle,  indicated  the 
decimal  place  of  the  digit  affected.  Among  the  first,  if  not  the  first,  to 
simplify  the  notation  of  decimals  by  the  use  of  the  decimal  point,  or 
comma,  was  John  Napier,  of  Scotland,  in  publications  of  1616  and  1617. 

Thus  it  is  seen  that  the  modern  fraction  and  the  modern  methods  of 
computing  with  fractions  are  the  result  of  many  stages  of  evolution, 
reaching  over  a  period  of  not  less  than  4000  years,  and  that  Asiatic, 
African,  and  European  nations  shared  in  effecting  this  evolution. 


CHAPTER  XV 

RADICALS  AND  GENERAL  EXPONENTS 

199.  All  numbers  of  elementary  algebra  can  be  placed  in 
one  of  two  groups,  —  real  numbers  or  imaginary  numbers. 

^•>  h  ~~  7->  4.675  are  real  numbers. 

V—  3,   —  2  V—  1,  §  V—  1  are  imaginary  numbers. 

Imaginary  numbers  or  expressions  are  numbers  which  in- 
volve the  square  roots  of  negative  numbers.  Real  numbers 
are.  rational  or  irrational.  Rational  numbers  are  positive  or 
negative  integers,  and  numbers  which  are  the  quotients  of 
such  integers,  such  as  9,  —  5,  %,  .125. 

Irrational  numbers  or  irrationals  are  numbers  that  are  real 
but  not  rational,  such  as  V2,  V9,  s/5. 

A  radical  is  a  root  of  any  algebraic  expression,  indicated  by 
the  use  of  a  radical  sign,  such  as  V9,  V6,  Va?  —  x  —  6. 

From  this  definition  it  is  evident  that  a  radical  may  be  a 
rational  or  an  irrational  number. 

The  integer  of  the  rational  sign  is  called  its  index.  Thus 
the  index  of  the  radical  sign  V  is  4.  If  no  index  appears, 
the  index  2  is  understood. 

A  number  by  which  a  radical  is  multiplied  is  called  its 
coefficient. 

Tbus,  in  7  V5,  7  is  the  coefficient  of  the  radical  VI. 

200.  The  value  of  an  irrational  number,  like  -\/2,  cannot  be 
expressed  exactly  in  Hindu-Arabic  numerals,  but  we  may 
represent  it  accurately  by  means  of  lines.  For  instance,  the 
diagonal  of  a  square  whose  side  is  1    is  equal  to  V2.     We 

178 


RADICALS  AND  GENERAL  EXPONENTS 


179 


cannot  find  V2  exactly,  but  we  can  draw  the  square  and  its 
diagonal. 


Fig.  34 


Likewise,  in  the  right  triangle  whose  legs  are  1  and  2,  the 
hypotenuse  is  Vl2  +  22  =  Vo. 

In  this  semicircle  whose  diameter  is  4,  the  perpendicular 
erected  at  the  distance  1  from  one  end  equals  V3,  as  is  shown 
in  geometry. 

FRACTIONS  AS  EXPONENTS 

201.  All  the  exponents  used  thus  far  have  been  positive 
integers.  We  defined  a5  as  a  •  a  •  a  •  a  •  a.  See  §  14.  That  is, 
the  exponent  5  indicates  that  a  is  taken  as  a  factor  five  times. 

What  is  the  meaning  of  a2  ?     It  would  be  absurd  to  say 

that  a2  signifies  a  taken  as  a  factor  one  half  times.  A  number 
can  be  taken  as  a  factor  only  a  whole  number  of  times;  for 
example,  three  times  or  ten  times. 

For  the  purpose  of  attaching  a  meaning  to  a¥,  we  stipulate 

that  we  shall  be  able  to  multiply  a?  by  a?  according  to  the 
same  rule  by  which  a2  is  multiplied  by  a2.  The  product  of  a2 
and  a2  is  found  by  adding  the  exponents.     That  is,  a-  •  a2  =  a4. 

If  we  add  the  exponents  in  the  multiplication  of  a2  by  a*  we 
obtain  afc+«or«ft     That  is, 


a2  •  a2  =  a. 


180  ELEMENTARY   ALGEBRA 

It  is  seen  that  a*  is  one  of  the  two  equal  factors  of  a,  or  the 
square  root  of  a. 

Hence  a?  is  another  way*  of  writing  the  square  root  of  a. 
i  _ 

That  is,  a  =  Va. 

In  the  same  way,  since 

8  S  8     i     3 

a2  •  a2  =  a2  *  *  =  a3,  , 
a?  is  another  way  of  writing  the  square  root  of  a3. 

That  is,  a%  =  Va3- 

Similarly,  since 

a*  •  a^  •  a*  =  a*  +  $  +  i  =  a2, 
it  follows  that  a*  =  Va2, 

and  ar*  =  Vxb,  etc. 

Thus  ^e  numerator  of  a  fractional  exponent  indicates  the 
power  of  the  base  and  the  denominator  indicates  the  root* 

202.  In  finding  the  value  of  8T,  we  may  take  the  cube  root 
of  8,  which  is  2,  and  square  2,  getting  4  as  a  result. 

Or  we  may  square  8,  which  gives  us  64,  and  then  take  the 
cube  root  of  64,  which  is  4. 

That  is,  8*  =  V8^  =  (V8)2. 

All  irrational  numbers  which  are  expressed  by  the  use  of 
radical  signs  can  be  expressed  by  the  use  of  fractional  expo- 
nents. In  fact,  the  simplification  of  expressions  can  be  effected 
more  easily  by  the  latter,  as  will  appear  later. 

Thus,  VWatb  =  16*aM  =  2  aM. 

i 
*  It  can  be  shown  that  Va  ora"  has  n  different  root  values,  but  in  ele- 
mentary algebra  only  one  of  these  values  is  usually  considered,  the  so-called 
principal  value.  If  a  is  &  positive  number,  then  the  principal  value  of  Va  is 
its  positive  root  ;  if  a  is  negative  and  the  index  n  is  odd,  there  is  no  positive 
root,  and  the  negative  root  is  taken  as  the  principal  root.  If  a  is  negative 
and  the  index  n  is  even,  then  all  the  roots  are  imaginary. 


RADICALS  AND  GENERAL  EXPONENTS         181 
ORAL   EXERCISES 

203.   Express  with  fractional  exponents  and  simplify : 
1.    -VcM>.  11.   4^aX 

2.  v^p;  12-  -2^^v. 

/s 13.    Qy/aW. 

3.  v  3  ran. 

4.  </2aiy. 

5.  V5  x^z2. 

6.  a-Vax3. 

7.  2V2^.  17.    -7v'-- 


14.  Vra  •  v  ra  •  Vra. 

15.  %y/w*  •  |Vto. 


16 


8.  V&.V&.  18  5V^ 

9.  24  a3^oy*.  19.  Va  +  6. 
10.  7^16  c2d3.  20.  3^/(a-&)2. 

EXERCISES 

204.    Write  with  radical  signs  : 

1.  a*.  9.  6  asy*. 

2.  6*.  10.  7  (3  m)*. 

3.  a&A  11.  a*&i 

4.  (7  a)*.  12.  3«*. 

5.  2(5»)*.  13.  (a?  +  y)*. 

6.  2  a*.  14.  4  (c  -  d)*. 

7.  (2  a)*.  15.  5a*(aj-y)*. 

8.  2  a^y*.  16.  a&». 


182  ELEMENTARY   ALGEBRA 

MEANING  OF  A  ZERO  EXPONENT 

205.  Thus  far,  no  meaning  has  been  assigned  to  a°.  Can 
we  not  discover  a  meaning  ?  By  the  law  of  exponents  in  di- 
vision (i.e.  subtracting  the  exponent  in  the  divisor  from  that 
in  the  dividend)  we  obtain  a3  -^  a3  =  a0.     But  when  a  is  not 


as 


zero,  a3  -7-  a3  =  —  =  1.     .  \  a0  =  1. 


a? 

Therefore  any  number  except  0  with  a  zero  exponent  is  equal 
to  1. 

MEANING  OF  A  NEGATIVE  EXPONENT 

206.    Reasoning  as  in  §  205,  we  obtain 

a5  -i-a?  =  or3,  also  a5  -=-  a8  =  —  =  — .     .*.  a~3  =  — 

a8      a3  a3 

Therefore,  any  quantity  with  a  negative  exponent  is  equal  to  1 

divided  by  that  quantity  with  the  exponent  positive. 

-    V         £ 

Further,  -«^L.  =    *  '        =  JL  =  ^1'  =  «* 

a-36-4c     1    1  _c_         a?c  c 

a3'  b*'        a3b* 
Therefore,  a  factor  may  be  moved  from  the  numerator  to  the 
denominator  of  a  fraction  provided  the  sign  of  its  exponent  be 
changed. 

Care  must  be  taken  to  ti'ansfer  only  factors  ;  in  - — i-= ,  a~2  is  not  a 

b 

factor  of    the    numerator.       -n—    is   not  equal    to       "*"    .      Since 

b  H  tfb 

1+1 
<r*  =  1 ,  it  follows  that  a'2  +  1  _l+a\ 


b  b  a26 

EXERCISES 


207.    Simplify: 

1.  86"*.  4.  7°.  16"*.  7.  (ft)"*. 

2.  64*.  5.  9*.  8.  (^j)i 

3.  125*.  6.  (-27)*.  9.  36*.  343*. 


RADICALS  AND  GENERAL  EXPONENTS         183 


10. 

c-sr1-*- (***)* 

• 

13. 

(- 

2)"2- 

(-3). 

11. 

12. 

9*  ■*■  (*)*• 

14. 
15. 

/25 
1 

2-i 

)-i.(x-y)°. 

1 
*3* 

16. 

31 
4-i" 

20. 

5° 
6°' 

24. 

4-2cr2 

17. 

f-Y-  — 

\5J     5-2 

21. 

2-*&. 

25. 

a"1  +  fir*. 

18. 

8° 

7  " 

22. 

3ar 

26. 

a-1  +  6-1 
a"1  -  6"1 

19. 

9 
3°" 

23. 

6a; 
y~3z 

REDUCTION  OF  RADICALS 

208.  A  radical  is  in  its  simplest  form  when  the  following 
three  conditions  are  all  satisfied  : 

(1)  There  are  no  fractions  under  the  radical  sign. 

VJ  is  not  in  its  simplest  form. 

(2)  The  number  under  the  radical  sign  contains  no  factor 
raised  to  a  power  equal  to  or  exceeding  the  index  of  the  radical 
sign. 

Va*  is  not  in  its  simplest  form,  because  the  exponent  4  exceeds  the 
index  3  of  the  radical  sign. 

(3)  In  a  radical  of  the  form  Va",  m  and  n  must  have  no 
common  factor  other  than  1. 

Vcfi  is  not  in  the  simplest  form,  because  2  and  4  have  the  common 
factor  2. 

Simplify  \/64  a862.     We  obtain 

^64  cPb*  =  2*  .  a%  •  b%  =  2*  a2b^  =  2  •  2^  •  a2b^  =  2  aV26. 


184  ELEMENTARY   ALGEBRA 


ORAL   EXERCISES 

209.    Simplify: 

i.  VI.  10     tjzT? 


2.    ^9.  V64s3 

3.  -t/w<*. 

4.    ^81. 


11. 


4/16a262 


5.  ^64^.  12.    ^ 

6.  </49^. 


(a  -  ft)2 

32cW° 
(a  +  2/)5 


7.  vi6«y. 


13.    V32^Y*. 


14.    V729xV2- 


</25  m2 
'     \49n2'  15.    Vab10c. 


e/4  a;2  fiTf, 

9y2'  *    27 

210.  1.    Simplify  V98. 

V98  =  V49  •  2,  wherein  49  is  the  largest  factor  of  98  which  is  a  perfect 
square.     V49  •  2  =  7\/2.    . ■ .  V98  =  7 V2. 

Again  5  v/54  =  5  v/27  •  2,  wherein  27  is  the  largest  factor  of  54  which  is 
a  perfect  cube. 

5^27^2  =5.3^2  =  16^2. 

2.    Simplify  Vf. 

Multiply  both  numerator  and  denominator  by  some  number  ichich  will 
make  the  denominator  a  perfect  square;  multiplying  by  3,  VI  =  Vf 
=  Vi  •  6,  wherein  1  is  the  largest  factor  of  6  which  is  a  square,  and  9  is 
used  for  its  denominator.   Since  VI  =  h  V&  '  8  =  JV^«    •  '•  Vf  =  1 V^' 

Again  7 ^M  =  7^iFT  =  ^rfWfl  =  ^^10. 

EXERCISES 

211.  Simplify: 

1.  V8.  3.    V20.  5.    V28. 

2.  V12.  4.    V24.  6.    V40. 


RADICALS  AND   GENERAL  EXPONENTS         185 


22.    2V54 

3 


7.  3V44.  27.  lOVlOOOG         47.    V^. 

8.  4V18.  28.  ^32.                  48.   3^|. 

9.  -6V27.  29.  3^/48.                49.    \/|. 

10.  5V48.  30.  7v80.                50.    Vf 

11.  2V56.  31.  ^224.               51.   3^S 

12.  7V121.  32.  Va?.                       52.    Ji. 

13.  -8V160.  33.  Vc^ 

14.  2V75.  34.  y/2tf.                53-   \ft' 

15.  3VI25.  35.  aVc^T 

16.  -9V300.  36.  bVtfy. 

17.  3V162.  37.  nWmhi. 

18.  2V320.  38.  5  VI 

19.  -fVl47.  39.  Vf.                        56-    |\J- 


54. 


55. 


ttVo3- 


1       1 

^a  +  2/' 


20.    |V108.  40.    V|. 

57. 


21.    V16.  41.    Vf  "  *x-y 


42-  VTV  58.    m»J-2--. 

23.  5^32.  43.  V5  m"n 

24.  7^56.  44.  |V^.  59.     ^'^ 

25.  3^88.  45.  vl-  r^~ 

26.  rfSGOl  46.  W,  6°-    ^-^V 


x-y 


ADDITION  AND  SUBTRACTION  OF  RADICALS 

212.    Similar  radicals  are  radicals  having  the  same  radical 
sign  and  the  same  number  under  the  radical  sign. 

iVa,  — 2Va,  b  va  have  the  same  radical  sign  V  ,  and  the  same  num- 
ber a  under  the  radical  sign.     Hence  they  are  similar. 


186  ELEMENTARY   ALGEBRA 

Add  2  V8,  3  VlO,  5  V50,  -  7  V40. 
Weobtain2v/8  +  3\/l0  +  5V50-    7V40 

=  4 V2  +  3 VlO  +  25V2  -  14 VlO  =  29 V2  -  llVlO. 

EXERCISES 

213.  Simplify: 

1.  a/3-2V3.  11.    V20+V45-V5. 

2.  5V2  +  3V2-V2.  12.    2V2-V18  +  3V50. 

3.  8V11-5V11.  13.   5V28  +  3V63-VII2. 

4.  9Va-2Va.  14.    VJ-2V2  +  V|. 

5.  aVb  +  bVb.  15.    Vo%+V63-V96. 

6.  2V2-3VJ.  16.   2V^6-Vo6c"2  +  V4o6. 

1: 5vf;^'        w-  >£+>^ 

9.    V|-V|.  18-    V(a  +  ft)3-Va2-62. 

10.    J«+v/*.  19-    V^+J^-J^-. 

20.    V(a^  —  y2)(ar  —  y)  — V4a^y2-(-4a;22/3. 

21.  Vl~6  +  \/54  +  \/l28.  23.    V8  a;  -  16  +  V27  a;  -  54. 

22.  ^5-^/7+^S-  24-    v/(a-3)2+V9a-27. 

MULTIPLICATION  OF  RADICALS  HAVING  THE  SAME  INDEX 

214.  Multiply  5Va  by  7V&  and  we  have 

5  •  7  •  Va  •  V&  =  35  Va6. 
Multiply  —  3V2  by  4  VlO  and  we  have 

-  3  •  4  •  V2  •  VlO  =  -  12  V20  =  -  24  V5. 

Rule.  Multiply  the  rational  factors  and  the  radical  factors, 
and  simplify. 

It  will  be  observed  that  the  multiplication  of  two  radicals, 
V#  and  s/y,  rests  on  the  identity  -\/xy  =  V#  •  V#. 


RADICALS  AND  GENERAL  EXPONENTS 


187 


ORAL   EXERCISES 

215.    Find  the  indicated  products : 

1.  V3  •  V27. 

2.  Vce  •  Var*. 

3.  V3-V5. 

4.  V3  •  V6. 

5.  V2a-  v^. 
6 
7 


V6x- VlOaf. 

-Vabc  •  —  -y/ae. 


8.    Va&V  •  Va?bc. 


11. 

V*-Vf. 

12. 

Vf-Vii. 

13. 

Vf.VJj. 

14. 

\lx-yjl 

15. 

Va2  •  Va"~26. 

16. 

-y/^y  •  "Van/"-1 

17. 

^2.^3. 

18. 

^I-^l- 

19. 

n-^l 

20. 

5^2-2^5. 

9.    V2a;2y  •  V4  an/. 
10.    #j&V  •  \/aW. 

216.   Multiply  V2  +  V3  -  V5  by  V2  -  V3  +  V5. 

V2+V3-V5 
V2-V3+V5 


2+: 
-3- 
-5 


^         1    +V15 

+vite+vi5 


-6  +2VI5 

Rule.     Multiply  each  term  of  one  polynomial  by  each  term 
of  the  other,  simplify,  and  combine. 

EXERCISES  _ 


217.   Multiply: 


1.  V6  -  V7  -  2  V£  by  V3. 

2.  V7+V3-V11  by  V5. 

3.  3V2-2V3+V5  by  V6. 


188  ELEMENTARY   ALGEBRA 

4.  V5-4V2+V10  by  V10. 

5.  V3  +  7V6-5VI2  by  V6. 

6.  3+V2  by  7-V3. 

7.  8  -  V6  by  V5  -  4. 

8.  3 V2  +  4V3  by  2V2  -  5V3. 

9.  2 V7  +  7 V2  by  2 V7  -  7V2. 

10.  5V3  +  3V5  by  5V3-3V5. 

11.  V6-V7+V8  by  V8+V7-V6. 

12.  2V5-3V6-4V7  by  V10-V5-V2. 
13.'  Vi+vT-V|  by  V5+V6-V7. 

14.  x y/y  +  -\/xy  —  yy/x  by  yfx  +  Vy. 

15.  -Z/2-J/3+J/I  by  ^4  +^/9-^2. 

DIVISION  OF  RADICALS  HAVING  THE  SAME  INDEX 

218.    1.   Divide  2  V2  by  V3. 

(a)   2V2--V3  =2\/|  =  fV6,  or 
(6)    2V2-,V3=^.y^=2V6  =  2V6. 
V3       V3         3         3 

Unless  the  division  of  the  quantities  under  the  radical  signs  gives  a 
whole  number,  the  second  method  is  preferable. 

In  (b)  we  express  the  division  in  the  form  of  a  fraction,  then  multiply 
both  numerator  and  denominator  by  some  number  which  will  make  the 
denominator  rational. 

2.   Divide  3  by  2  +  V3. 

3^-(2+V3)  =      3        •  2-V3=6-3a/3  =  6_3V3 
2  +  V3      2-V3        4~3 
Again,  (3V2 +4\/3)-i-(3V2  -  4V3) 
_3v/2+4v/3      3v/2  +  4V3_  18  +  24  a/6+48  _  66  +  24V6  _     33  +  12V6 
~  3^2-4  V3  '  3V2+4V3~        18-48  -30  16 


RADICALS  AND  GENERAL  EXPONENTS         189 

If  the  denominator  is  of  the  form  Va  ±  Vb,  multiply  both  terms  of  the 
fraction  by  Va  T  Vb.    This  is  called  rationalizing  the  denominator. 

To  find  the  numerical  value  of  2  V2  -f-  V3  in  its  present  form  requires  the 
extracting  of  two  square  roots  and  a  long  division.  It  is  much  less  work 
to  rationalize  the  denominator,  then  in  the  resulting  expression,  |V6, 
find  a/6,  and  take  §  of  it. 

EXERCISES 


8.  Vcc-l-y  by  s/x  —  y. 

9.  5  by  V5  +  1. 

10.  3  V7  by  V7  -  4. 

11.  V2+V3by  V5+V61 

12.  V3  +  V5  by  V2  -  V6. 

13.  Vi0-V2by  V3-V5. 

14.  vT5  -  V3  by  V5  +  V6. 

15.  VH  -  VI2  by  Vli  +  VI2. 

16.  2 V2  +  3 V3  by  2V2  -  3 V2. 

17.  2  V3  -  3  V2  by  2  V3  +  3  V2. 

18.  a  V6  +  b  Va  by  ay/b  —  6  Va. 

19.  a  —  Va  +  6  by  Va  +  b. 

20.  y/x  -\-  y  —  y/x  —  y  by  VaT+  y-f-Va  —  y. 

RADICAL  EQUATIONS 


219 

.  Divide : 

l. 

V2  by  V5. 

2. 

1  by  V3. 

3. 

24  by  V6. 

4. 

2  by  \/2. 

5. 

36  by  V48. 

6. 

6  by  2  -  V3. 

7. 

a  by  Va~-+-  b. 

220.    1.   Solve  Va  -  2  =  5. 

Square  both  sides,  x  —  2  =  25  ;  a;  =  27.     Am.  27. 

OAecA; :  V27  -  2  =  5 ;  5  =  5. 


190  ELEMENTARY  ALGEBRA 


2.   Solve  Vx  +  5  +  -Vx  —  1  =  3. 


Transpose,  Vx  +  5  =  3  —  Vx  —  1 . 

Square  both  sides,  x  +  5  =9  —  6Vx  —  1  +  x  —  1. 

Transpose  and  combine,  6-v/x  —1=3. 


Vx-1=$. 

Square  both  sides,  x—  \  =  \\  x  =  J.     Ans.  f. 

Check :  VfTb  +  VpTi  _  3. 

V*t  +  V\  =  3. 

f  +  £=3. 

3  =  3. 

221.  Rule.  If  the  equation  contains  but  one  radical,  trans- 
pose so  that  it  will  stand  alone. 

If  the  equation  coyitains  two  radicals,  transpose  so  that  there  is 
one  radical  on  each  side  and  so  that  the  rational  terms  are  com- 
bined on  one  side. 

Raise  both  sides  to  a  power  corresponding  to  the  degree  of  the 
radicals.     Solve  the  resulting  equation. 

In  squaring  both  sides  of  an  equation,  a  value  of  x  may  be  obtained 
which  will  not  satisfy  the  original  equation.  Such  a  value  is  called  an 
extraneous  root.     For  instance,  squaring  both  sides  of  the  equation 

Vx  +  5=1—  Vx, 

there  results  x  +  5  =  1  —  2Vx  +  x. 

Transpose  and  simplify,  2  =  —  y/x. 

Square  both  sides,  4  =  x. 

Substitute  in  Vx  +5  =  1—  Vx,  3=1  —  2,  which  is  absurd. 

Hence  x  —  4  is  not  a  root  of  the  original  equation,  but  an  extraneous 
root.  In  fact,  there  is  no  value  of  x  that  will  satisfy  Vx  +  6  =  1  —  Vx. 
In  other  words,  this  equation  has  no  roots. 

Only  by  substitution  in  the  original  equation  involving  radicals  can  we 
tell  whether  a  value  obtained  for  x  is  a  root  of  that  equation  or  whether 
it  is  an  extraneous  root. 


RADICALS  AND  GENERAL  EXPONENTS         191 


222 

E2 

.    Solve  and  check : 
Vx  -  5  =  2. 

IERCISBS 

l. 

6.   a;  =  5  —  Vcc2  —  a;  —  6. 

2. 

3  Va  -  2  =  1. 

Vx  +  5  =  2  Vx  +  2. 

7.    ^4a;-16=2. 

3. 

8.    8  -  Vx  -  1  =  5  +  Vx  +  2. 

4. 

3Va-3=V72. 

9.   x  =  13  -  V.t2  -  13. 

5. 

z-3  =  -Va2-6. 

10.   3  —  Vz  =  Va  +  3. 

11.    Va2-7a; 

V*  +  1  _  Va7  -  3 
Vx  —  2      Vx  +  4 

+  6  =  Vz2  -  a;  -  12. 

12. 

15     Vx  13—'" 

Vx  +  3 

13 

Va;  +  17  =  a;  —  3. 

K 

16.    Vx      2           D        -0. 

14.    V—  x—  5  —  x  =  7.  Vx  —  2 


CHAPTER   XVI 

REVIEW   OF   THE  ESSENTIALS   OF   ALGEBRA 
DEFINITIONS 

223.  Define  the  following : 

Coefficient  Monomial  Linear  equation 

Exponent  Binomial  Quadratic  equation 

Term  Polynomial  Boot  of  an  equation 
Factoring 

EXERCISES   ON   PARENTHESES 

224.  Remove  the  parentheses  and  simplify  : 

1.  (5  _  6)  -  (7  -  8)  +  (5  -  6  -  7)  -  (-  5  -  4  +  6). 

2.  ll_(3-[4-5])  +  [4-(3-5)+(4-l)]. 

3.  4(3  -  2)  +  4(3  h-  2)  -  4  -=-  (4  -  2). 

4.  (4  a  +  b)  -  (8  b  -  c)  -  (a  -  7  &). 

5.  Ax2-7y2-  [7x2-6x  +  8y2~]-  [5x-3y~]. 

6.  3^-5^-^2a^-4^-9?/2|  -\1  rf-bxy]. 

7.  a  -  [b  +  (c  +  2  d)  -  (-  c  -  d)]  -  [2  a  -  3  6]. 

8.  -(a  -  cTd)  +  (2  a  -  3  b  +  4  c)  -  (4  a  -  [5  b  -  6  c]). 

Group  all  terms  with  a  single  letter  in  a  parenthesis  with 
+  before  it,  and  all  terms  with  two  different  letters  in  a 
parenthesis  with  —  before  it : 

9.  a—  6-f-4c  —  a&  +  2ac  —  4  6c  +  a?b. 

10.  — x-\-5y  —  4:Z  +  xy  —  3xz-\-7yz. 

11.  2^-5?/2  +  322-2a;z  +  2?/z  +  4a;y. 

12.  cde  —  led  —  3de+f—  d  +  2c  -3e. 

192 


REVIEW  OF  THE  ESSENTIALS  OF  ALGEBRA     193 

EXERCISES  IN  ADDITION,  SUBTRACTION,  MULTIPLICATION, 
AND   DIVISION 

225.    Simplify: 

1.  2x  +  (x  —  l)2.  6.  2m(ffl-4)-m2. 

2.  lt+(2t  +  iy.  7.  3(x  +  y)-2(x-y). 

3.  4 1  +  (2 «  -  f).2  8.  (x  +  yf  +(x-  yf. 

4.  4y-(2y  +  l)2.  9.  (a-b)2+{a-by. 

5.  (x  —  yy  +  ixy.  10.  a(a-26)  +  (a+&)(a-6). 

Find  the  values  of  the  following  at  sight : 

11.  43,  34,  41,  l4,  2°,  8°,  25*. 

12.  a?a3,  a4a,  4  cV,  ccV,  y5y5. 

a     o^   4  <r     g        5     6        o 
a9    610   5  c6 

4a263   12  a354   15  a5?/5    75  a;2yz 
6  a3b  '    4  a&2  '  25  x5?/ '  15  x^y^z 

Simplify : 

15.  (a  +  by+(a-by.  18.  (a  +  2  6)2  -(a  -f  &)(<*-&). 

16.  (a-2  6)2+(a  +  6)2-  19.  (2  a  -  6)2  +  (a  -  &)2. 

17.  (a  +  &-2c)2.  20.  (a-&)2-(a+2&)(a-2  6). 

Multiply  orally : 

21.  (*  +  3)(a?  +  9).  25.  (6_5)(6-4). 

22.  (y  +  7)^  +  4).  26.  <2-z)(3-z). 

23.  (7-z)(5-z).  27.  (z2  +  7)(z2  +  5). 

24.  (a-6)(a  +  8).  28.  (ay  +  l)(ay  +  4). 

Perform  the  indicated  divisions  : 

29.  (f  +  2tf-y-2)+(y  +  l). 

30.  (^  +  21y  +  342)-i-(y  +  6). 

31.  (4x8-5a!*  +  20a?  +  50a!2  +  2a;-12)-*-(a!-2a?+6aH-5). 


194  ELEMENTARY  ALGEBRA 

EXERCISES   IN    FACTORING 

226.   Factor: 

1.  10  xf  -  15  aty»-  8-    3  a;2  (a;  -  1)  -  (1  -  a?). 

2.  (2x  +  y)a  +  (2x  +  y)b.  9.   3m(a2  +  62)  -  2  n  (62  +  a2). 

3.  5  »(o  +  6)  -  3  a;  (a  +  6).  10.    (m  -  k)2  -  (m  -  to). 

4.  ic6  —  5  xh  —  3  x*  +  3  a*  11.   ab  —  ax  +  by  —  xy. 

5.  a,-6 -5  a* +  ^-5  x3.  12.   a3  -  a26  +  ab2  -  b3. 

6.  (a-b)a2-(a-b)b2.  13.   y2  +  10?/  +  25. 

7.  (a;2  -  1)  m  +  (a;2  -  1)  n.  14.   3?  + 8  a; +  16. 

15.  a2b2  +  12  a&  +  36. 

16.  a2  +  b2  +  c2  —  2  ab  -  2  ac  +  2  6c. 

17.  m2  —  49.  32.    62  -  12  &  —  45. 

18.  64  -x2.  33.   c2  +  14c-32. 

19.  81aJ*-25y«.  34.   a;2 +  10  a; -75. 

20.  (a-6)2-c2.  35.    a? -10  a; -75. 

21.  x2-(5x  +  y)2.  36.   x2  -  7  23  -  8  ?/2. 

22.  a2  +  2a&  +  &2-c2.  37.    a2  +  15  x2y  +  14  xhf\ 

23.  a;2  -5  x  +  4.  38.   70  a2  +  17  a  +  1. 

24.  a2  +  5  x  +  4.  39.   2  a;2 +  9  a; +  9. 

25.  y*  +  6y-16.  40.   2-y-6y2. 

26.  2/2  —  6 y  —  16.  41.   6-m-15m2. 

27.  a?-7a;  +  10.  42.   a3  +  8  b3. 

28.  x2  +  7a;  +  10.  43.    a3  -  8  b\ 

29.  z2  +  3x-10.  44.    8  ar»  -  27  */*. 

30.  x2-3a;-10.  45.    8  a?  +  27  y3. 

31.  a2  +  7a-18.  46.   a4-!. 


REVIEW  OF   THE  ESSENTIALS  OF  ALGEBRA    195 


EXERCISES   ON   FRACTIONS 

227.    l.   Showthat  ±^  =  ^  =  -^  =  -±-^  =  +  ?. 
+  b      -6  +6         -b         6 

2.   Showthat  +a-  +a- 


3. 


—6+6  +6  6 

?  1  ? 


x  —  y      y-x'    (x-y)(z-x)      (x  —  y)(x-z) 
1  ?  1  ? 


(x-y)2      (y-xy     (x-y)(x~z)      (y-x)(z-x) 
Reduce  each  fraction  to  its  lowest  terms : 

5  6Q  +  y) 

'  30  +  y) 

6  5(x  +  y)3 
'   20(x  +  yy 

7. 
8. 


2a-26 
a2  +  ab 


9 

3-2c 

9 -4c2 

10. 

6a;  +  aa; 

wa2  —  w&2 

11 

5  x2  +  5  xy 

x2  —  y2 

io 

x2  +  x  —  6 

-  a6  (a  -  2)3 

Multiply : 

13-  TJ  by  dhr  15-  -tH*  by 


y3         16a;2  2a;      J  ?/2  -  a;2 

14    15 gjy  .     34z^  a;y2  - y3  b    x*-xy2 

17 z2      '  5a^  x2  +  a*/     J       5y2 

Divide : 

17.  aLzi|  by  -  f  ^  2T  ^  •  20.  (a2 -a;2)  by  a  +  a; 

,0m  —  n,       c  +  w  _,     a4  —  4  6, 

18. by  — - 21.   by  (a2  — 2  6) 


a-»-27 

x2-\-2x  + 

1 

c  +  n 

m  +  n 

3m 

c  —  n         ra  +  n  a  +  6 

3m  __     a;2  —  5  i 

m  —  w  a;2  —  2  a;  —  35  ~J    x  +  5 


,_       6  ra3     ,        3m  __     x2  —  5 a;  +  6    ,      a;  —  3 

19-  -x — —%  by  z — r  22-  ^ — ttz-^^  by 


196  ELEMENTARY   ALGEBRA 

Perform    the    indicated    additions    and     subtractions     of 
fractions : 

23.  l-£=l.  26.    -1 L_. 

a3         a5  (x-1)3      (1-xy 

24.  JL  +  **-k*.  27.   ^-  +  _3_. 
2y      3y     8y  »  — y     x  +  y 

M.-J i—  M.-S i,'. 

(x  —  y)3      (y  —  x)2  a  —  3      a  —  4 

Reduce  each  expression  to  an  equivalent  fraction : 

««    o      .  1— 5  a;  __  3  m  — 4 

29.  3aH 32.    m 

4  3  —  m 

30.  2y  +  5x~3y-  33.    x2  +  ay  +  y*  +  -£_  . 

4  x  —  y 

2  a2  m1 

31.  3  a —  •  34.    x2  —  xy  +  y* 2 

a  +  b  x+y 

EXERCISES   IN   LINEAR  EQUATIONS 

228.    Solve: 
1.   8  x  +  19  =  11  -5  x  +  13.         3.   |««i2  — a 

2-   \y  +  7=-  iy  +  8.  4.  ■  |a?-2-|aj=:4aj. 

5.  (16  cc  +  5) (9  aj  +  31)  =  (4  x  +  12)(36  cc  +  10). 

6.  x  (x  +  1)  +  x  (x  +  2)  =  (a  +  3)(2  a  +  4). 

7.  mx  =  nx  +  3.  a;  +  4  _  - 

8.  ay  +  by  =  c.  x  —  5 

9.  2  ay  -  2  by  =  16  +  2  a.  15.    5  -  -  =  3. 


10    aa;  +  6a;  —  ca;  =  d. 


SB 

16      ft"1-2 

11.  mx  —  nx  —  a;  =  ra  +  n.  ±0*    ^  ,  2  —  5 

12.  (a>  +  a)»=»  {«-&)*.  g  _  1  =  g  _  3 

13.  (a;  — 4)2  — (a;— 5)2=4a;+5.  a;  —  4      a;  — 2 


REVIEW  OF   THE  ESSENTIALS  OF  ALGEBRA    197 

is.  x~1  ■  x~3 .  26   \x  +  y  =  35> 

2x-l      1x-2  '     U-y  =  18. 


-M—  =  JtM-  -2.  9~     |2  as  +  2/  =  74, 

Sl  +  ^=L  28.    (3-  +  22/  =  16, 


19. 
20. 


4  x  +  5  y  =  33. 

z-1      a,-  +  l  29     jy  =  3a-5, 
3x-5^      2  Ua;-22/-5  =  0. 

»  — 4  3  |?_2/_i 

a?  +  4  _  a-  +  1  30-      3      5~    ' 
2x           2     *  l»-y  =  l 

y2  +  y  +  3  =  y  +  1  81     |-3a;  +  8?/  =  5, 
2w-H             2  lo?  +  w  =  2. 


2y  +  l  2  \x  +  y 

25     2^+5  =  ^±5>  32      f3a;  +  cy  =  3c2j 

2a;  +  l      a?  +  7  13  a; -cw  =  c2. 


EXERCISES   IN   QUADRATIC  EQUATIONS 

229.    Solve: 

12.  loa  +  -  =  11. 

2.  a;2  —  5  x  +  5  =  0.  % 

3.  a;2 -5  a;  =  -4.                          ,„  1      _      1 

13.  a;  +  -  =  5+-. 

4.  (3aj-2)(aj-l)=14.  a;             5 


5.    (3a;-2)(aj-l)=200. 


14.    _^  +  ^_=8. 


6.  (2x-l)2=-8a;.  l  +  »     1 

7.  a?  -  2  aa:  =  b\  ,  -     a?  -  1         d 

15. 


X 


8.  a;2  -  (a  -  1)*  +  a  =  0.  c         x  +  1 

9.  a;2  +  wkb  +  n  =  0.  16     a-'  +  l  ,  fl  +  2 
10.  j*c2  +  qx  +  r  =  0.  a;  —  1      a  —  2 

ii.  -JL^^+2.  i7.  \3x  =  2y> 

x  —  1      x2  —  1  |  a*/  =  54. 


198 


ELEMENTARY  ALGEBRA 


18. 


19. 


20. 


21. 


2x  =  Sy, 
{  -3y2  =  24-2x2. 

x2  +  y2  =  ll, 

3  x  —  y  =  —  1. 

y2  =  4  x, 
{x  +  y=15. 

J  *    y 


22. 


23. 


24. 


1-5=2. 
x     y 


x7  +  y2  =  25, 
x2  -  y2  =  7. 

x2  =  16  y, 
x  =  2y. 


PROBLEMS 

230.    1.    What  number  added  to  the  numerator  and  denomi- 
nator of  ^  will  give  a  fraction  equal  to  f  ? 

2.  A  train  runs  300  miles  in  a  certain  time.  If  the  train 
were  to  run  10  miles  an  hour  faster,  it  would  take  it  1  hour 
less  to  travel  the  same  distance.     Find  the  time  and  rate. 

3.  The  length  of  the  rim  of  one  of  the  hind  wheels  of  a 
carriage  exceeds  the  length  of  the  rim  of  one  of  the  front 
wheels  by  3  ft.  The  front  wheel  makes  the  same  number  of 
revolutions  in  running  200  ft.  that  the  hind  wheel  makes  in 
running  260  ft.     Find  the  length  of  the  rim  of  each  wheel. 

4.  A  farmer  found  that  the  supply  of  feed  for  his  cows 
would  last  only  20  weeks.  He  therefore  sold  50  cows,  and  his 
supply  lasted  30  weeks.     How  many  cows  had  he  at  first  ? 

5.  Two  trains  start  together  and  run  in  opposite  directions, 
one  at  the  rate  of  47  miles  an  hour,  the  other  at  the  rate  of  34 
miles  an  hour.  After  how  many  hours  will  they  be  1053 
miles  apart  ? 

6.  The  area  of  a  rectangle  is  1850  sq.  in.  Its  length 
exceeds  its  width  by  13  in.     How  long  is  each  side  ? 


REVIEW  OF   THE  ESSENTIALS  OF  ALGEBRA    199 

7.  A  farmer  expected  to  receive  $5.52  for  his  eggs.  He 
broke  6  eggs,  but  by  charging  2  ^  a  dozen  more  for  the  remain- 
ing eggs,  he  received  the  desired  amount.  How  many  dozen 
eggs  had  he  originally  ? 

8.  Weighed  on  a  defective  balance  a  body  appears  to 
weigh  P  pounds  when  the  true  weight  is  a  -f-  bP  pounds,  where 
a  and  b  are  numbers  that  are  the  same  for  all  weights.  If  the 
balance  shows  10  pounds  for  a  real  weight  of  11.2  pounds,  and 
30  pounds  for  a  real  weight  of  33.2  pounds,  find  a  and  b. 
What  will  the  balance  show  for  a  real  weight  of  40  pounds  ? 

9.  According  to  tradition,  the  following  problem  was 
assigned  by  Euclid  of  Alexandria  to  his  pupils,  about  3  cen- 
turies b.c.  :  A  mule  and  a  donkey  were  going  to  market  laden 
with  wheat.  The  mule  said  to  the  donkey,  "  If  you  were  to 
give  me  1  measure,  I  would  carry  twice  as  much  as  you,  if  I 
were  to  give  you  1  measure,  our  burdens  would  be  equal." 
What  was  the  burden  of  each  ? 

10.  A  room  shaped  as  in  Fig.  37  has  a  floor 
area  of  320  sq.  ft.  If  the  lengths  marked  a 
are  all  alike,  and  b  is  22  ft.,  find  a. 

11.  To  pass  from  one  corner  of  a  rectangu- 
lar park  to  the  opposite  corner  I  must  go  700 
feet,   if   I   go  around  the   sides ;    if    I    walk 
diagonally  across,  I  save  200  feet.     What  are  the  dimensions 
of  the  park  ? 

12.  It  takes  a  cyclist  18  minutes  longer  to  go  a  distance  of 
27  miles  from  A  to  B  than  it  takes  to  return  by  the  same 
route.  If  he  rides  up  hill  6  miles  an  hour  and  down  hill  15 
miles  an  hour,  how  many  miles  of  his  trip  going  is  up-hill 
travel,  and  how  many  miles  is  down  hill  ? 

13.  A  dealer  sells  bicycles  at  a  20  %  profit.  A  rival  dealer 
obtains  the  same  kind  of  bicycles  $  3  cheaper  and  sells  them 
$3  cheaper,  thereby  realizing  a  profit  of  21|  °J0-  What  price 
does  the  first  dealer  pay  for  his  bicycles  ? 


.. 


200  ELEMENTARY   ALGEBRA 

14.  If  speculum  metal  contains  67  %  of  copper  and  33  %  of 
tin  (by  weight)  and  gunmetal  contains  90  %  of  copper  and 
10  °J0  of  tin,  how  many  pounds  of  gunmetal  should  be  melted 
with  300  pounds  of  speculum  metal,  to  obtain  an  alloy  in 
which  there  is  5  times  as  much  copper  as  tin  ? 

15.  Water  in  freezing  expands  10  <f0  of  its  volume.  How 
much  water  when  frozen  will  fill  a  5-gallon  freezer  ? 

16.  In  a  price  list,  the  cost  of  sewer  pipe,  per  foot  of  length, 
is  given  by  the  formula  c  =  .4  d2  +  14,  where  d  is  the  diame- 
ter of  the  pipe  in  inches  and  c  is  the  cost  in  cents.  What  will 
400  ft.  of  15  in.  pipe  cost  ? 

PROBLEMS   ON   THE  LEVER 

231.  1.  What  force  must  be  applied  by  the  hand  to  balance 
or  raise  a  known  weight  W=  748  pounds  ? 


i         i         i         i         i 


H 

Fig.  38  L~ J 

By  careful  measurement  it  has  been  ascertained  that  the  force  multi- 
plied by  the  force  arm  of  the  lever  is  equal  to  the  weight  multiplied  by  the 
weight  arm.  Notice  that  the  weight  arm  is  measured  from  the  point  of 
application  of  the  weight  to  the  prop  or  fulcrum,  A,  and  the  force  arm  is 
measured  from  the  point  of  application  of  the  force  to  the  prop.  In  this 
case  /  =  4,  w  =  2. 

Let  F  =  the  required  force  in  pounds,  then 

4  F  -  2  x  748. 
F  =  374  pounds,  the  answer. 

2.  How  large  a  weight  can  be  raised  with  a  force  of  500  lb., 
if  the  force  arm  is  7.5  ft.  and  the  weight  arm  .5  ft.  ?  Draw 
a  figure. 

3.  The  force  arm  is  16  times  greater  than  the  weight  arm. 
What  force  will  balance  a  weight  of  732  pounds  ? 


REVIEW  OF   THE  ESSENTIALS  OF  ALGEBRA    201 

4.  The  force  arm  is  2  ft.  longer  than  the  weight  arm.  How 
long  is  the  weight  arm,  if  a  force  of  12  pounds  balances  a 
weight  of  35  pounds  ? 

5.  A  lever  is  75  in.  long.  Where  must  the  prop  be  placed 
in  order  that  a  force  of  2  pounds  at  one  end  may  move  5 
pounds  at  the  other  end  ? 

6.  Two  men  carry  a  load  of  212  pounds  on  a  pole  between 
them.  If  the  load  is  3.5  ft.  from  one  man  and  5.4  ft.  from  the 
other,  how  many  pounds  does  each  man  carry  ? 

7.  A  man  weighing  170  pounds  has  a  crowbar  6  ft.  long. 
What  pressure  can  he  exert  toward  moving  a  rock,  if  the  prop 
is  3  in.  from  the  lower  end  of  the  crowbar  ? 

8.  On  an  untrue  balance  a  weight  of  13  oz.  appears  to 
weigh  13.2  oz.  If  the  beam  is  6  in.  long,  and  the  error  is  due 
to  a  displacement  of  the  fulcrum,  how  much  longer  is  one  arm 
than  the  other  ? 

9.  A  team  is  hitched  to  a  doubletree  4  ft.  long.  At  what 
point  must  the   doubletree   be 


puiiiu    111  us  l    one    uuuuiei/iet;     uv  n              n              n              n 

attached  to   the   plow,  so  that  c     ■    H          JJ     ^    H     n          n 

one  horse  will  pull  14  times  as  a          w          li          g    |i    I 

mnn.h  as  t.hfi  nt.hp.r  ?  <U          "                  — * 


D 


10.    Each      of       two      horses  l^J  Plow-Beam 

hitched  to  the  doubletree  A  in  F     39 

Fig.  39  pulls  4  as  much  as   a 

horse  at  B.     At  what  point  should  a  plow  beam  be  attached  to 
the  doubletree  evener  CD,  which  is  5  ft.  long  ? 

PROBLEMS    ON   FALLING   BODIES 

232.    Bodies  falling  from  a  state  of  rest  obey  laws  expressed  by  the 
following  formulas  :  v  _  32.2 1 

8  =  16.1 12, 
v  =  V64.4  s, 
where  v  means  the  velocity,  t  the  time  in  seconds,  s  the  space  or  distance 
(in  feet)  through  which  the  body  falls. 


202  ELEMENTARY  ALGEBRA 

1.  Express  each  of  the  three  formulas  in  words. 

2.  Eliminate  t  from  the  first  two  formulas,  and  thereby  de- 
rive the  third  formula. 

4 

3.  Eliminate  s  from  the  last  two  formulas,  and  thereby  de- 
rive the  first. 

4.  Eliminate  v  from  the  first  and  third,  and  thereby  de- 
rive the  second. 

5.  A  stone  dropped  into  a  mine  shaft  is  heard  to  strike 
bottom  after  5  seconds.  Neglecting  the  time  it  takes  sound 
to  travel,  estimate  the  depth  of  the  shaft. 

6.  With  what  velocity  will  a  body  strike  the  bottom  of  a 
mining  shaft  1000  ft.  deep  ? 

7.  A  stone  is  heard  to  strike  the  bottom  of  a  mine  shaft 
after  ten  seconds,  sound  traveling,  in  this  instance,  at  a  rate 
of  1126  ft.  a  second.     What  is  the  depth  of  the  shaft  ? 


INDEX 


Numbers  refer  to  sections 


Absolute,  term,  166. 

value,  29. 
Addition,  algebraic,  26,  28,  29. 

checking,  45. 

fractions,  193. 

monomials,  41. 

polynomials,  43. 

radicals,  212. 
Ahmes,  165. 
Algebraic  expression,  2. 

evaluation  of,  4,  23. 
Area,  circle,  134. 

rectangle,  63. 

trapezoid,  134,  174. 

triangle,  134. 
Arithmetical  operations,   order 

21. 
Average,  35. 
Axis,  z-axis,  90;  y-axis,  90. 

Bank  discount,  100. 
Base,  13,  14. 
Binomial,  12. 
Braces,  17. 
Brackets,  17. 

Cardan,  183. 
Circle,  134,  182. 

area  of,  134. 

length  of,  8,  134. 
Checking,  addition,  45. 

division,  137. 

equations,  51. 

factoring,  113. 

multiplication,  60. 

subtraction,  45. 
Coefficient,  10. 

of  radical,  199. 
Concrete  mixture,  83. 
Coordinates,  147. 


of, 


Degree  of  an  equation,  130. 
Descartes,  165. 
Dissimilar  terms,  41. 
Division,  36. 

by  zero,  5,  137. 

checking,  36,  74,  137. 

law  of  exponents,  65. 

of  fractions,  70,  186. 

of  monomials,  65. 

of   polynomials   by   binomials, 
76. 

of  polynomials  by  monomials, 
74.  ) 

of  polynomials  by  polynomials, 
135. 

of  radicals,  218. 

with  remainders,  79. 


Elimination,  151,  162. 

by  addition  or  subtraction,  152. 

by  substitution,  154. 
Equation,  5. 

complete  quadratic,  166,  169. 

containing  fractions,   98,    156, 
160,  195. 

cubic,  130. 

degree  of,  130. 

double  root,  132. 

fractional  coefficients,  98. 

general  quadratic,  175. 

graphical    solution    of    quad- 
ratic, 179. 

graph  of,  89,  90. 

history  of  quadratic,  183. 

incomplete  quadratic,   167. 

involving  parentheses,  54,  62. 

linear,  90,  130. 

literal,  51. 

plotting,  89,  90. 

quadratic,  130,  166. 

quartic,  130. 
203 


204 


INDEX 


Equation,  radical,  220. 

root  of,  51,  132. 

solution  of,  6,  51. 
Equations,  inconsistent,  146. 

literal,  158. 

one  linear,  one  quadratic,  180. 

simultaneous  linear,   147,   151, 
162. 
Exponent,  13,  14. 

fractional,  201. 

law  for  division,  65. 

law  for  multiplication,  56. 

negative,  206. 

zero,  205. 
Extraneous  roots,  221. 

Factor,  9. 

highest  common  factor,  188. 

prime,  112. 

rationalizing,  218. 
Factoring,  107. 

checking,  113. 
Formulas,  distance,  8,  100. 

interest,  3,  100. 
Fractions,  65,  184. 

addition,  193. 

complex,  187. 

division,  70,  186. 

history  of,  198. 

multiplication,  68,  186. 

multiplication  by  integer,  69. 

reduction,  65,  185. 

signs  of  terms,  69. 

subtraction,  193. 
Function,  89. 

Gauss,  183. 

Graphic  representation,  84. 
Graphs,    construction   and   use   of 
graphs,  97. 
identical,  147. 
of  linear  equations,  90.  / 
of  quadratic  equations, ,  179. 
of    simultaneous   linear   equa- 
tions, 145. 
parallel,  147. 

practical    application    of,    94, 
149. 

Highest  common  factor,  188. 
Hyperbola,  182. 


Imaginaries,  199. 
Imaginary  unit,  178. 
Income  tax,  100. 
Inconsistent  equations,  146. 
Index,  of  radical,  199. 

of  root,  15. 
Interest,  100. 

Least  common  denominator,  193. 
Least  common  multiple,  190. 

Magic  squares,  49. 
Monomial,  12. 

addition,  41. 

division,  65. 

multiplication,  56. 

subtraction,  41. 
Multiplication,  33. 

checking,  60. 

fractions,  68,  186. 

law  of  exponents,  56. 

monomials,  56. 

numbers,  33. 

polynomials,  58,  60,  135. 

radicals,  214. 

Negative  numbers,  24,  26. 

Two  uses  of  +  and  -,  28,  39. 
Newton,  frontispiece. 
Number,  absolute  value  of,  29. 

integral,  112. 

unknown,  5,  6. 
Numbers,  addition  of,  26,  28,  29. 

division  of,  36. 

imaginary,  178,  199. 

irrational,  110,  199,  200. 

multiplication  of,  33. 

negative,  24,  26. 

positive,  24,  26. 

rational,  199. 

real,  199. 

subtraction,  31. 

Origin,  90. 

Parabola,  180. 
Parentheses,  17. 

equations  involving,  53,  62. 

insertion  of,  48. 

removal  of,  46. 
Pi,  134. 


INDEX 


205 


Plotting,  84,  90,  145,  146. 
Polynomials,  12. 

addition,  43. 

division,  74,  135. 

multiplication,  58,  60,  135. 

subtraction,  43. 
Portraits,  Descartes,  page  140. 

Euler,  page  40. 

Newton,  frontispiece. 

Vieta,  page  110. 
Positive  numbers,  24,  26. 

uses  of  +  and  — ,  28,  39. 
Power,  13. 

cube,  13. 

fourth,  13. 

square,  13. 
Problems,  algebraic  solution,  6. 

arithmetical  solution,  6. 

bank  discount,  100. 

distance,  100. 

falling  bodies,  232. 

income  tax,  100. 

interest,  100. 

lever,  231. 

concrete  mixture,  83. 

taxes,  100. 
Products,  special,  101. 
Proportion,  81. 

direct,  83. 

inverse,  83. 

Quadratic  equation,  130,  166. 
complete,  166,  169. 
general,  175. 
graph  of,  179. 
history  of,  183. 
incomplete,  166,  167. 
solved     by     completing 

square,  171. 
solved  by  factoring,  130. 
solved  by  formula,  176. 

Radical,  199. 

coefficient  of,  199. 

equation,  220. 

index  of,  199. 

sign,  15. 
Radicals,  199. 

addition,  212. 

division,  218. 

multiplication,  214. 


the 


Radicals,  rationalization,  218. 

reduction  of,  208. 

similar,  212. 

simplify,  111. 

subtraction,  212. 
Ratio,  81. 

Rational  number,  199. 
Recorde,  165. 
Rectangle,  area  of,  63. 
Root,  cube,  15,  109. 

fourth,  15. 

index,  15 

of  an  equation,  51. 

principal,  108,  201. 

square,  15,  108,  139,  142. 
Roots,  extraneous,  221. 

imaginary,  178. 

Sign,  change  of,  in  fractions,  69. 

not  equal  to,  113,  note. 

of  equality,  1. 

of  operation,  39. 

of  quality,  39. 

radical,  15. 
Similar,  radicals,  212. 

terms,  41. 
Simultaneous  equations,   151,   162, 

180. 
Square  root,  15,  108,  139. 

of  fractions,  143. 

of  numbers,  142 
Subtraction,  26,  28,  31. 

checking,  32,  45. 

fractions,  193. 

monomials,  41. 

polynomials,  43. 

radicals,  212. 

Term,  11. 

Terms,  dissimilar,  41. 

similar,  41. 
'Transposition,  51. 
Trapezoid,  134. 

area  of,  134,  174. 
Triangle,  altitude,  134. 

area  of,  134. 

hypothenuse,  134. 

leg,  134. 
Trinomial,  12. 

Unknown,  5,  6. 


206 


INDEX 


Variables,  88,  89. 
Vieta,  165. 
Vinculum,  17. 

Volume  of  rectangular  solid,  64. 
of  cube,  64. 

Whetstone  of  Witte,  165. 


X-axis,  90. 
F-axis,  90. 

Zero,  division  by,  137. 
exponent,  205. 
operations  witn,  130. 


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